Median of Two Sorted Arrays
10 Prerequisite LeetCode Problems
“Median of Two Sorted Arrays” requires an understanding of binary search, divideandconquer algorithm design, and array manipulation. Here are 10 simpler problems to build the necessary skills:
“Two Sum” (LeetCode problem #1): A classic problem that involves array manipulation and hashing. It is a basic problem that anyone should start with before moving onto more complex problems.
“Merge Sorted Array” (LeetCode problem #88): This problem is easier but requires knowledge about merging sorted arrays, which is a prerequisite for understanding the median of two sorted arrays.
“Binary Search” (LeetCode problem #704): It is essential to understand the binary search algorithm before attempting the “Median of Two Sorted Arrays” problem. This problem provides a good way to practice implementing a binary search.
“Search Insert Position” (LeetCode problem #35): This problem applies binary search in a slightly more complex scenario, which is a good practice after understanding the basic concept.
“Find First and Last Position of Element in Sorted Array” (LeetCode problem #34): This problem requires binary search to solve. It can help you to understand how to modify binary search for different scenarios.
“Valid Palindrome” (LeetCode problem #125): A problem that is not directly related, but which encourages you to think about handling edge cases and testing your implementation.
“Remove Duplicates from Sorted Array” (LeetCode problem #26): This helps you to understand how to manipulate arrays and work with sorted arrays.
“Find Minimum in Rotated Sorted Array” (LeetCode problem #153): This problem introduces a more complex application of binary search in the context of sorted arrays.
“Squares of a Sorted Array” (LeetCode problem #977): It helps to understand how to process sorted arrays and use the twopointer technique, which is useful for many arrayrelated problems.
“Remove Element” (LeetCode problem #27): This problem can help you get a better understanding of array manipulations.


Problem Classification
This problem belongs to searching and sorting.
Let’s break down the components:
Sorted Arrays: We are given two sorted arrays,
nums1
andnums2
, of sizem
andn
respectively. This implies that we might need to leverage the fact that these arrays are already sorted to optimize our solution.Return the Median: The task is to find the median of the two sorted arrays. This is the primary goal.
Time Complexity: The overall runtime complexity needs to be O(log(m+n)). This implies that the problem is likely asking for a binary search solution or a similar logarithmictime algorithm.
This can be further classified into a subcategory within searching and sorting: “Merge and Find”. The classic way to find a median from a sorted list involves merging the two lists and then finding the middle. However, the logarithmic time complexity constraint indicates we need to use a more efficient method.
This is also a classic example of the divideandconquer strategy which forms the basis of binary search algorithms. In divideandconquer problems, we typically break the problem into smaller subproblems of the same type, solve those subproblems, and combine the results.
This is a combination of understanding array data structures, knowing searching algorithms (particularly binary search), and having a grip on some basic statistical measures (like the median).
Understand the importance of leveraging properties of the input data (in this case, sorted arrays) to optimize solutions.
This can be classified as a Statistics/Data Analysis problem.
Because the problem deals with two sorted arrays of numbers (which can be viewed as ordered data sets), and the task is to determine the median value, which is a common statistical measure that indicates the middle point of a data set. Therefore, the main theme of this problem is statistics and data analysis, thus fitting into the Statistics/Data Analysis category.
Clarification Questions
Questions to ask for clarity:
What should be the output when both input arrays are empty? The problem specifies that
m
andn
, the sizes ofnums1
andnums2
, can be zero, but it doesn’t specify the output for this case.In what format should the median be returned? Should it always be in decimal format or should it match the format of the input numbers? For example, if all the numbers are integers and the median logically is an integer, should it be returned as an integer or a float?
Is it guaranteed that the inputs will always be valid? For example, are the arrays guaranteed to be sorted, and will they only contain numbers? This could influence whether we need to add error checking and data validation in our solution.
What is the expected behavior if there are multiple possible medians? For example, in some definitions, if a set of numbers is [1, 2, 3, 4], both 2 and 3 can be considered medians. Should the algorithm favor the lower or higher median, or can it return either? In this case, the problem seems to suggest that the arithmetic mean should be returned (i.e., (2+3)/2 in this example), but it would be good to confirm this.
Are negative numbers or zeros allowed in the arrays? The constraints indicate that negative numbers are indeed allowed, but it would be good to confirm this.
The goal of asking these questions is to ensure that you have a thorough understanding of the problem and the edge cases before you start working on a solution. It’s important to understand the constraints and expectations fully to design the most efficient and accurate solution possible.
Problem Analysis and Key Insights
The problem statement provides several insights:
Presorted Arrays: The arrays
nums1
andnums2
are already sorted. This is a valuable insight because it means we can use methods that leverage sorted data, such as binary search, to achieve logarithmic time complexity.Median Calculation: The problem requires us to find the median of two sorted arrays. This suggests a merge operation, but a typical merge operation of two arrays would result in linear time complexity. So, we have to find a more efficient way to get the median.
Logarithmic Time Complexity: The requirement to achieve a time complexity of O(log(m+n)) is significant. It rules out methods that would require us to merge the two arrays fully (as that would be linear time complexity). This requirement points us towards a binary search type solution or other divideandconquer methods, which typically have logarithmic time complexity.
Constraints: The constraints of the problem (such as array sizes and the range of the numbers) suggest that we don’t have to worry about overflow problems in our calculations. Also, we can expect large inputs, so an efficient solution is necessary.
Edge Cases: Identifying potential edge cases can provide key insights. For instance, cases where one or both of the arrays are empty or when the total number of elements in both arrays is even will need special handling in our solution.
Remembering these insights can guide you towards an efficient solution and also help avoid common pitfalls during implementation.
Problem Boundary
The scope of this problem encompasses several aspects:
Understanding of Arrays: You need to understand how to work with arrays, a fundamental data structure in computer science. This includes accessing elements, understanding array lengths, and potentially slicing arrays.
Search Algorithms: The problem specifically suggests that the solution should have a logarithmic time complexity, indicating that a binary search or a similar divideandconquer strategy might be needed.
Statistics Concept: The problem requires finding the median, which is a statistical measure. Therefore, a basic understanding of statistics, particularly the concept of a median, is required.
Efficient Problem Solving: The constraints of the problem (sizes of arrays and the range of numbers) suggest that efficiency is important. The solution needs to be carefully designed to handle potentially large arrays without exceeding time limits.
Edge Case Handling: There are several potential edge cases, such as when the arrays are of different lengths, when one or both arrays are empty, or when the total number of elements in both arrays is even. The solution needs to handle these correctly.
In terms of implementation, the scope will depend on the programming language being used. It will likely involve defining a function that takes the two arrays as input and returns the median as a float.
The scope of a problem can often be larger than it initially appears, and understanding the full scope can help guide your problemsolving approach.
The boundary of this problem can be defined based on its requirements and constraints. Here’s how we can establish these boundaries:
Input Constraints: The problem statement provides clear boundaries in terms of input size. The length of both arrays combined (
m + n
) must be at least 1 and can go up to 2000. Each array can have a minimum length of 0 and a maximum length of 1000. The numbers in the arrays can range from 10^6 to 10^6. Any solution must handle inputs at these extremes.Algorithm Complexity: The problem also sets a clear boundary on the time complexity of the algorithm, which should be O(log(m+n)). This effectively rules out solutions that would operate in linear time or worse, and points us towards binary search or other divideandconquer strategies.
Output Requirements: The output needs to be the median of the numbers in the two input arrays, represented as a decimal number. If the combined length of the arrays is even, the median is the average of the two middle numbers.
Edge Cases: There are a few edge cases that establish the boundary of the problem. For example, when one or both input arrays are empty, or when the combined length of the arrays is even, the solution needs to handle these situations correctly.
These elements, combined, define the boundary of the problem. Any solution must be able to handle all possible inputs within these constraints and produce the correct output as specified.
Distilling the Problem to Its Core Elements
This problem is based upon the fundamental concept of binary search and array partitioning. Binary search is a divideandconquer algorithm used to find a particular element in a sorted array. The array partitioning is about dividing the array at a certain point such that certain conditions hold true. Here, it is applied to find the median of two sorted arrays by making partitions such that all elements on the left side are smaller than those on the right.
In simplest terms, imagine you have two sorted lists of numbers, and you want to find the middle number when both lists are combined together. You can’t just combine them because it might take a lot of time for large lists. So, you need to find a faster way.
The core problem we are trying to solve is to find the median of two sorted arrays in logarithmic time complexity. To simplify: How can we efficiently find the middle number from two sorted lists of numbers without merging them?
Breaking down the problem into key components, we have:
 Two sorted arrays, nums1 and nums2.
 A requirement to find the median value between these two arrays.
 A constraint to achieve it in O(log(m+n)) time complexity.
To solve this problem, at a minimum, we need to perform these operations:
 Initiate a binary search on the smaller array to find a suitable partition point.
 Calculate the partition point in the other array using the partition point in the smaller array.
 Compare elements at the partition points to adjust the partition in the next iteration or calculate the median if the correct partition is found.
 If the total length of the arrays is odd, the median is the maximum of the left partition. If it’s even, the median is the average of the maximum of the left partition and the minimum of the right partition.
Visual Model of the Problem
Visualizing this problem can be helpful to understand the partitioning concept and how we determine the median. Here is a way to visualize it:
Let’s take an example where we have two sorted arrays:
nums1 = [1, 3, 8]
nums2 = [2, 7, 9]
Combine them without changing the order:
[1, 3, 8] and [2, 7, 9]
We want to partition each array in such a way that the total number of elements in both left partitions is equal to the total number of elements in both right partitions, and every element in the left partitions is less than or equal to every element in the right partitions.
Possible partitions of the arrays:
[1, 3  8] and [2  7, 9]
In this partition, the left side elements are 1, 3, 2
and the right side elements are 8, 7, 9
. But here max(3, 2) > min(8, 7)
, which means we have to adjust our partition.
Adjusting the partition:
[1  3, 8] and [2, 7  9]
Now, the left side elements are 1, 2, 7
and the right side elements are 3, 8, 9
. Here max(1, 7) <= min(3, 8)
, which is what we want.
The median then depends on the total number of elements. If it’s odd, it’s the maximum of the left partition, otherwise it’s the average of the maximum of the left partition and the minimum of the right partition. In this case, we have six elements in total, so the median is (max(1, 7) + min(3, 8)) / 2 = (7 + 3) / 2 = 5
.
This visual representation can provide a better understanding of how we are manipulating the two arrays to find the median without merging them.
Problem Restatement
You are given two lists of numbers. Both lists are sorted in nondecreasing order, meaning each number in a list is not less than the one before it. The lengths of the lists can vary and can be anywhere from 0 to 1000. You need to find the middle value from the combined set of numbers from both lists. This middle value is referred to as the median.
If the total count of numbers is odd, the median is the number exactly in the middle. If the count is even, the median is the average of the two middle numbers. However, instead of physically combining the lists, which can take significant time for large lists, you are required to find a faster method, specifically with a time complexity of O(log(m+n)), where m and n are the sizes of the two lists respectively.
Essentially, the problem is about efficiently finding the middle value from two sorted lists without merging them.
Abstract Representation of the Problem
This problem belongs to a class of problems that can be described abstractly in terms of order statistics, sorted sequences, and binary search.
Sorted Sequences: The problem deals with two sorted sequences of numbers. The sequences could be abstractly represented as
sequenceA
andsequenceB
. Both sequences are in nondecreasing order, meaning each element in a sequence is not less than the one before it.Order Statistics: The problem requires finding the median, which is a type of order statistic. Order statistics are about the properties of data related to its order or rank. In this case, the median is the middle value or average of two middle values in the ordered set of all elements from both sequences.
Binary Search: The problem needs to be solved with an algorithm having a time complexity of O(log(m+n)). This immediately suggests binary search, a technique that halves the search space in each step, leading to logarithmic time complexity.
Partitioning: The crux of the solution involves finding an appropriate partition point in both sequences such that all elements on the left are less than or equal to all elements on the right, and the total count of elements on both sides are equal (or the left side has one more if the total count of numbers is odd).
Thus, in an abstract representation, this problem involves applying binary search for efficient partitioning of two sorted sequences to identify the correct order statistic (median).
Terminology
There are a few specialized terms and technical concepts that are crucial to understanding this problem and its solution.
Median: The median is a measure of central tendency. In a sorted list of numbers, the median is the middle value if the list’s size is odd. If the size is even, it’s the average of the two middle numbers. In this problem, the median helps us understand where to partition our arrays  on either side of the median.
Sorted Arrays: An array is a data structure that stores a collection of items. An array is sorted if its elements are in a specific order. In this problem, we’re dealing with arrays sorted in nondecreasing order, which enables the use of binary search.
Binary Search: Binary search is an efficient algorithm for finding an item in a sorted list. It works by repeatedly dividing in half the portion of the list that could contain the item until you’ve narrowed down the possible locations to just one. In this problem, binary search is used on the smaller array to find an appropriate partition point that divides both arrays into two halves such that all elements in the left halves are less than or equal to all elements in the right halves.
Time Complexity: Time complexity is a computational concept that describes the amount of time an algorithm takes to run as a function of the size of the input to the program. In this problem, we’re asked to find a solution with a time complexity of O(log(m+n)), which means the running time increases logarithmically in relation to the input size.
Partitioning: Partitioning an array means dividing it into two sections based on a condition. In this problem, we want to partition the arrays in such a way that all elements on the left are less than or equal to all elements on the right, and the number of elements on both sides are equal or the left side has one more.
Understanding these terms and concepts is crucial for comprehending the problem and developing an efficient solution.
Problem Simplification and Explanation
Let’s break down this problem:
We have two sorted lists of numbers, and we want to find the middle number when both lists are combined. Now, we could just merge the lists and find the middle, but that could be slow for very large lists. So, we’re tasked with finding a faster way.
The key concepts here are:
Median: The median is like the balance point in a list of numbers. When all the numbers are lined up, the median is the one right in the middle. If there are even numbers of items, the median is the average of the two middle numbers.
Sorted Arrays: This is like having two rows of numbered blocks, each arranged from smallest to biggest.
Binary Search: Imagine you’re searching for a word in a dictionary, but instead of going page by page, you open the dictionary in the middle and see if the word you’re looking for comes before or after the page you’re on. If it comes before, you now only have to search the first half of the dictionary. You repeat this process, halving the section of the dictionary each time, until you find the word. This is binary search  it’s a way to find something quickly in a sorted list.
Partitioning: Imagine you have a group of people, and you want to split them into two groups so that everyone in one group is shorter than everyone in the other. This would be a partition of the people based on height.
Applying this to our problem: We want to split our two lists in such a way that every number in the left parts of both lists is smaller than every number in the right parts, and the number of elements on both sides are equal or the left side has one more.
Think of this like having two rows of blocks and trying to draw a single line that divides the blocks into two groups, with all smaller blocks to the left of the line and all larger blocks to the right, balancing the quantities on either side.
The challenge here is to draw this line quickly (in O(log(m+n)) time), and that’s where we use binary search. We apply binary search on the smaller list to find where to draw this line. This will help us figure out where to draw the line in the larger list as well, leading us to find the median quickly.
Constraints
This problem has a few characteristics and constraints that we can exploit to find an efficient solution:
Sorted Arrays: The problem provides us with sorted arrays, which is a significant advantage. With sorted arrays, we can make certain assumptions that wouldn’t be valid with unsorted data. For instance, we can perform a binary search, which takes advantage of the sorted nature of arrays to find elements efficiently.
Median and Partitioning: The problem asks for the median, which is a form of partitioning. If we can partition the two arrays correctly, the median will be immediately accessible. This gives us a clear goal for what kind of partitioning we’re aiming for.
Logarithmic Time Complexity: The problem specifically asks for a solution with O(log(m+n)) time complexity, which suggests binary search or some form of divideandconquer strategy, taking advantage of the sorted nature of the arrays.
Numerical Ranges: The constraints on the length of the arrays (0 <= m,n <= 1000) and the range of the numbers in the arrays (10^6 <= nums[i] <= 10^6) are quite broad, but the fact that these ranges are finite and not excessively large means that our algorithm won’t have to deal with exceedingly large numbers or incredibly long arrays, which helps ensure that our solution will be practical and efficient.
Overall, these characteristics point towards a binary search solution where we search for the correct partition point in the two arrays that separates smaller and larger numbers, helping us find the median efficiently.
Analyzing the constraints gives us important insights that guide the approach to solving this problem:
Array Size: The size of the input arrays is limited (0 <= m,n <= 1000). This is important because it limits the amount of data the algorithm will need to process, giving us the confidence that a solution with a logarithmic time complexity will be efficient even for the maximumsized inputs.
Value Range: The range of values in the arrays is from 10^6 to 10^6. This is not an exceedingly large range, which means we won’t have to handle excessively large or small numbers. Also, the fact that the range is defined gives us an upper and lower limit to work with.
Sorted Arrays: The arrays are sorted. This is a crucial constraint, which enables the use of binary search or other efficient search algorithms that depend on the input being sorted.
Median and Logarithmic Time Complexity: The problem requires finding the median in O(log(m+n)) time complexity. The fact that the solution must operate within logarithmic time pushes us towards efficient search techniques like binary search or divideandconquer algorithms.
Combined Array Length: The total length of the combined arrays is reasonably small (1 <= m+n <= 2000). This means the problem of finding the median can be solved efficiently with the right algorithm.
From these insights, we can infer that an efficient solution should likely use a binary search or divideandconquer approach to identify the correct partition in the arrays, from which the median can be found.
Case Analysis
Here are additional examples or test cases that cover a wide range of the input space, including edge and boundary conditions:
Small Input Size: This will help us test the basic functionality of the solution. For example, if the input array is [1, 3], and we’re trying to find the position of 3, the expected output would be 1. In this case, the array has the minimum size of 2 elements, yet it’s sorted and the target exists.
Large Input Size: This is a stress test for the algorithm. Let’s say we have an array of size 10^6 with elements from 1 to 10^6 in ascending order, and we’re trying to find the position of 500,000. The expected output would be 499,999 (since the index starts from 0). This tests the efficiency of the binary search algorithm on large input sizes.
Target at the End: In this case, the target is at the end of the array. If the input array is [1, 2, 3, 4, 5] and we’re trying to find the position of 5, the expected output would be 4. This case checks if our binary search algorithm correctly handles situations where the target is at the edge of the array.
Target at the Beginning: Similar to the previous case, but the target is at the beginning of the array. For example, if the input array is [1, 2, 3, 4, 5] and we’re trying to find the position of 1, the expected output would be 0.
Target Not in the Array: The target value doesn’t exist in the array. For example, if the input array is [1, 2, 3, 4, 5] and we’re trying to find the position of 6, the expected output would be 1 or some indicator that the target doesn’t exist. This test checks if our algorithm handles the case of the target not being present in the array.
Empty Array: This is a boundary case where the input array is empty. In this case, regardless of what target value we’re looking for, the output should be 1 or an indicator that the array is empty.
Single Element Array: The array contains only one element. If the input array is [1] and we’re trying to find the position of 1, the expected output would be 0. This test verifies if our algorithm handles the case of the smallest nonempty array.
These examples test the binary search algorithm against various edge cases, helping to ensure its robustness and correctness.
Visualizing these test cases can be an effective way to understand the problem and solution space better.
Small Input Size: Draw an array with just two elements and mark the position of the target.
[1, 3] 0 1 Target: 3
Large Input Size: Illustrate a large array with dots in the middle indicating that there are many more elements, and mark the position of the target.
[1, 2, …, 500000, …, 10^6] 0 1 ^ ^ Target: 500000
Target at the End: Draw an array and mark the position of the target at the end.
[1, 2, 3, 4, 5] 0 1 2 3 4 Target: 5
Target at the Beginning: Draw an array and mark the position of the target at the beginning.
[1, 2, 3, 4, 5] 0 1 2 3 4 Target: 1
Target Not in the Array: Draw an array and indicate that the target is not present.
[1, 2, 3, 4, 5] 0 1 2 3 4 Target: 6 (not present)
Empty Array: An empty array can be represented by an empty set of brackets.
[] Target: any number (not present)
Single Element Array: Draw an array with one element and mark the position of the target.
[1] 0 Target: 1
In all of these diagrams, the numbers below the arrays represent indices, starting from 0. For the ‘Target Not in the Array’, ‘Empty Array’, and ‘Single Element Array’ cases, the output would be 1, 1, and 0 respectively.
Analyzing different cases provides several key insights:
Small Input Size: When the input size is small, most algorithms, even inefficient ones, will be able to find the target relatively quickly. However, for larger datasets, an efficient algorithm like binary search shines, as it keeps reducing the problem space by half with each iteration.
Large Input Size: Binary search performs very well on large input sizes due to its logarithmic time complexity. This is in contrast to a linear search, which would take significantly longer as the input size increases.
Target at the End or Beginning: Even though the target is at the extreme ends of the array, binary search doesn’t need extra steps. It continues to reduce the search space until it finds the target. This shows that binary search doesn’t favor the location of the target.
Target Not in the Array: Binary search will continue to reduce the search space until it realizes that the target is not present. Understanding this behavior is important for implementing the exit conditions of the algorithm correctly.
Empty Array and Single Element Array: These cases represent the minimum possible input sizes. They help us understand the behavior of the algorithm when the input size is at its smallest, thereby informing us about any necessary base or edge cases to handle in the algorithm.
By analyzing these cases, we understand the strength of binary search in dealing with large inputs and how it doesn’t favor any particular location of the target. We also understand how it behaves when the target is not present and when the input size is at its minimum.
Identify Invariant
In the context of this problem, the invariant condition  the condition that remains true throughout the execution of the algorithm  is the sorted nature of the two input arrays and the definition of median.
Here’s how these invariant conditions apply:
Sorted Arrays: The input arrays are assumed to be sorted in nondecreasing order. This invariant allows us to apply a binary search and also lets us assume that if a partition point is valid in one array, the corresponding partition point in the other array will also be valid. This is critical for the algorithm to work correctly.
Definition of Median: The median of a sorted list is defined as the middle element if the length of the list is odd, or the average of the two middle elements if the length is even. This definition remains constant and is used to calculate the final result.
These invariant conditions guide the problemsolving approach and are assumed to hold true at each step of the algorithm. The algorithm is designed based on these invariants, and it uses them to make certain assumptions. Therefore, if these invariants were to be violated, the algorithm would not function correctly.
Identify Loop Invariant
A loop invariant is a condition that is initially true and remains true after each iteration of a loop. It’s a concept used in formal correctness proofs for algorithms.
In this problem, the loop is the while loop where the binary search is performed. The loop invariant here is the search space where the correct partition point could exist. Before the loop starts (initialization), we establish the search space by setting start
to 0 and end
to the length of the shorter array.
During the loop (maintenance), we adjust the search space by updating start
and end
based on the comparison of elements at the partition points. If maxLeftX
> minRightY
, it means we’ve gone too far to the right in array nums1
, and we should move to the left, i.e., we decrease end
to partitionX  1
. If maxLeftY
> minRightX
, it means we’ve gone too far to the left in array nums1
, and we should move to the right, i.e., we increase start
to partitionX + 1
. This ensures that our search space keeps shrinking with each iteration, but it always includes the correct partition point.
At the end of the loop (termination), when start
> end
, the loop breaks and at this point, we have found the correct partitions that satisfy the condition maxLeftX <= minRightY
and maxLeftY <= minRightX
, which indicates that all elements on the left are less than or equal to all elements on the right. This maintains the sorted nature of the arrays and lets us compute the median based on the partition points.
Thus, the loop invariant in this problem is the fact that the search space always contains the correct partition point.
For this problem, the invariant and the loop invariant are not the same. They refer to different properties that hold true in different contexts of the algorithm:
Invariant: The two arrays being sorted is an invariant in this problem. This property is crucial for the binary search algorithm to work correctly. It is not specifically associated with the loop, but rather with the overall problem context.
Loop Invariant: In the context of the binary search loop in the solution, the loop invariant is the property that the search space always contains the correct partition point. This property is initially true and remains true after each iteration of the loop. It is directly associated with the functioning and correctness of the loop.
Therefore, while both the invariant and loop invariant are important for the correct operation of the algorithm, they are not the same for this problem.
Identification of Applicable Theoretical Concepts
The problem revolves around some key mathematical and algorithmic concepts that help simplify the problem and make it more manageable:
Median Calculation: Finding a median is a standard statistical operation. The median is the middle point in an ordered list of numbers. It divides the dataset into two halves with an equal number of elements or with the left half having one more element if the total count is odd. Understanding this property is crucial to the problem.
Binary Search: Binary search is a powerful algorithmic concept used in computer science for efficiently finding an element’s position in a sorted array. It repeatedly divides the array into two halves until the target element is found. This problem is an excellent candidate for binary search due to the sorted nature of the arrays and the requirement for a logarithmic time complexity solution.
Partitioning Concept: In this problem, we aim to partition the two arrays such that elements on the left are all less than or equal to the elements on the right, with the number of elements in both halves being equal (or the left half having one more element if the total count is odd). This can be seen as an application of the ‘Order Statistics’ concept in computer science.
Divide and Conquer Strategy: Divide and Conquer is an algorithmic paradigm that breaks problems into smaller subproblems, solves the subproblems independently, and combines their solutions to solve the original problem. In this case, the binary search approach we employ to find the partition point essentially divides the problem into smaller subproblems, making the overall problem more manageable.
Understanding and applying these mathematical and algorithmic concepts can significantly simplify the problem and guide us towards an efficient solution.
Simple Explanation
Imagine you have two lines of people arranged from shortest to tallest, and you want to find the person who is in the middle when both lines are combined. One way to do this would be to merge the two lines into one big line and then find the person in the middle. However, this would take a lot of time as you’d have to rearrange all the people.
Here’s a quicker way to do it: If we could draw a line in each group such that all the people on the left of the line in both groups are shorter than all the people on the right of the line, and the total number of people on the left and right are about the same, then the person at the boundary of the left and right would be our middle person.
So, our challenge is to draw these lines quickly. And how do we do that? Well, we can do something similar to looking up a word in a dictionary  instead of starting from the first page, we start in the middle, and depending on whether the word we’re looking for comes before or after the page we’re on, we either look in the first half or the second half of the dictionary. This is called binary search, and it helps us find the line in the people groups quickly, leading us to the middle person.
Problem Breakdown and Solution Methodology
Let’s break down the process of solving this problem.
To find the median of the two sorted arrays, we want to partition both arrays such that the elements on the left are all less than the elements on the right and that there are an equal number of elements on both sides (or the left side can have one more element if the total count of elements is odd). This allows the median to be directly found from the partition points.
Think of it as playing a game of “guess the number,” where the number lies between 1 and 100. You wouldn’t start by guessing 1, then 2, then 3, and so on. Instead, you would start with 50. If you learn the number is lower, you go to 25, and if it’s higher, you go to 75. This is the concept of binary search.
We use a similar binary search strategy on the smaller of the two arrays to find the partition points. By iteratively adjusting the partition points based on whether the maximum element on the left side is greater than the minimum element on the right side, we can find the correct partitions that satisfy the above conditions.
Let’s walk through an example:
Consider two arrays, nums1 = [1, 12, 15, 26, 38] and nums2 = [2, 13, 17, 30, 45]. We want to find the median of these two arrays.
Step 1: Identify the smaller array, which is nums1 in this case. We’ll perform the binary search on this array.
Step 2: We initialize two pointers, start and end, to the beginning and end of nums1 respectively.
Step 3: We calculate the midpoint of nums1, and using it, we calculate the partition point in nums2 such that the total elements on the left side of the partition points in both arrays is equal to the total elements on the right side (or one less if the total count is odd).
Step 4: Check if the maximum element on the left side of both arrays is less than or equal to the minimum element on the right side of both arrays. If so, we have found the correct partition. The maximum of the elements on the left side of the partition in both arrays will be the maximum element in the combined left partition, and the minimum of the elements on the right side of the partition in both arrays will be the minimum element in the combined right partition. If the total number of elements is even, the median is the average of the maximum element on the left and the minimum element on the right. If it’s odd, the median is the maximum element on the left.
Step 5: If the maximum element on the left is greater than the minimum element on the right, we need to move the partition point in nums1 to the left to decrease the maximum element on the left. Hence, we adjust the end pointer to mid  1 and repeat the process from step 3.
Step 6: Similarly, if the maximum element on the left is less than the minimum element on the right, we move the partition point in nums1 to the right by setting the start pointer to mid + 1 and repeat the process from step 3.
This process is repeated until we find the correct partition, which will give us the median. This approach ensures that we are able to find the median efficiently in O(log(min(m,n))) time, where m and n are the sizes of the two arrays. This efficiency is due to the binary search strategy we use, which cuts down the search space by half in each iteration.
If the elements in the arrays were to change or the size of the arrays was to increase, this approach would still hold as it is not dependent on the actual elements in the arrays, only their order. Hence, it works effectively for any two sorted arrays.
Inference of ProblemSolving Approach from the Problem Statement
Median: This is a statistical term that denotes the middle value in an ordered sequence. If the sequence has an even number of observations, the median is the average of the two middle numbers. In this problem, our task is to find the median of the merged array.
Sorted Arrays: This indicates that the input arrays are already sorted in ascending order. This is important because it simplifies our task and enables us to use binary search to find the median. If the arrays weren’t sorted, we would first need to sort them which would add to the time complexity.
O(log (m+n)) Complexity: This is a time complexity constraint indicating that the solution needs to be more efficient than a linear scan of the arrays. This points us towards using a divide and conquer approach, such as binary search, which has a logarithmic time complexity.
Binary Search: This is an efficient algorithm for finding an item from a sorted list. It works by repeatedly dividing in half the portion of the list that could contain the item, until you’ve narrowed down the possible locations to just one. In this problem, we use binary search to efficiently find the correct partition point.
Partition: This is a term that refers to the division of an array into two segments. In this problem, we aim to partition both arrays such that all elements on the left are less than those on the right, and there are an equal number of elements on both sides, or one more on the left if the total number of elements is odd.
These terms and concepts guide our approach to solving the problem using binary search to efficiently find the partition points that will yield the median of the two sorted arrays.
Creating a table or diagram to visualize this problem can be helpful in understanding the properties and the algorithm’s workings.
Sorted Arrays: List the elements of the two arrays in ascending order, each array on a separate line.
nums1: 1, 12, 15, 26, 38 nums2: 2, 13, 17, 30, 45
Partitions: Draw vertical lines to represent possible partition points. For instance, a partition after the 3rd element in nums1 and the 2nd element in nums2:
nums1: 1, 12, 15  26, 38 nums2: 2, 13  17, 30, 45
Binary Search: Illustrate the iterative binary search process with a decision tree. For example, if maxLeft1 > minRight2, move the partition in nums1 to the left:
mid1=3


 
mid1=2 mid1=4


 
mid1=1 mid1=3
Each node in the tree represents the midpoint in nums1 at each step. Left child indicates moving the partition point to the left, and right child indicates moving to the right.
 Median Calculation: Once you’ve found the correct partitions, list the elements on each side and identify the max on the left and the min on the right. In this case, the maxLeft = 15 and minRight = 17, so the median is the average, (15+17)/2 = 16.
This visual representation will make it clear how binary search iteratively adjusts the partition points to find the median. It shows how the algorithm narrows down the search space and balances the partitions to satisfy the conditions of a median.
The following points from the problem statement guide us toward the use of a binary search strategy:
Sorted Arrays: The arrays given in the problem are already sorted. Binary search is an algorithm that is specifically designed for efficient searching in sorted data structures.
Find Median: We are required to find the median of the two arrays combined. In a sorted array, the median is a positional statistic  it doesn’t depend on the exact values of the elements, but their position in the sorted order. This allows us to use binary search to find the correct position for the median.
Time Complexity Requirement: The problem statement specifies a time complexity of O(log(m+n)). Binary search has a time complexity of O(log(n)) because it halves the search space at each step, which aligns with the required time complexity.
Given these cues, we can infer that binary search is an apt algorithmic strategy to solve this problem efficiently.
Simple Explanation of the Proof
Let’s take a step back and discuss the thought process behind this algorithm. The goal is to find the median of two sorted arrays combined, without actually merging them. Let’s keep in mind the main property of the median: in a sorted set, the median separates the lower half from the upper half of the data.
Given two sorted arrays nums1 and nums2, let’s start by making an initial guess that the median is in the middle of nums1. This is where we start our binary search.
We make a partition in nums1 such that all elements on the left are less than those on the right. We do the same in nums2 based on our guess in nums1, so that the total number of elements on the left and right of the combined partition is equal (or left has one more if the total number of elements is odd).
Now, we need to check if we have found the correct spot for the median:
 We look at the elements near the partition: maxLeft1 and maxLeft2 (the largest elements on the left side of the partition in nums1 and nums2 respectively), and minRight1 and minRight2 (the smallest elements on the right side of the partition in nums1 and nums2 respectively).
 If maxLeft1 <= minRight2 and maxLeft2 <= minRight1, then we have found the correct partition. All elements on the left side are less than or equal to all elements on the right side. The median is the average of max(maxLeft1, maxLeft2) and min(minRight1, minRight2) if the total number of elements is even, or max(maxLeft1, maxLeft2) if it’s odd.
If we have not found the correct partition, we adjust our binary search in nums1:
 If maxLeft1 > minRight2, this means that the partition in nums1 is too far to the right. We need to move it to the left, so we adjust our binary search to look at the left half of nums1.
 If maxLeft2 > minRight1, this means that the partition in nums1 is too far to the left. We need to move it to the right, so we adjust our binary search to look at the right half of nums1.
We repeat this process until we find the correct partition.
The reason why this works is that binary search efficiently narrows down the search space for the partition in nums1. Because the arrays are sorted, we can ensure that adjusting the partition maintains the sorted order in the combined array. The invariant here is that the elements on the left side of the partition are always less than or equal to the elements on the right side. By maintaining this invariant, we ensure that once the correct partition is found, we have the median of the two arrays.
Stepwise Refinement
Let’s break down the approach into more granular steps, identify independent parts and repeatable patterns.
Stepwise Refinement:
a. Input Verification: Verify if either of the input arrays, nums1 or nums2, is empty. If an array is empty, the median is simply the median of the other nonempty array.
b. Initialization: Initialize two pointers, min_index and max_index, to keep track of the beginning and end of the smaller array (nums1 if nums1.length <= nums2.length else nums2). Start binary search on the smaller array.
c. Partitioning: For each midpoint in the binary search of the smaller array, calculate the partition index in the larger array using the formula
(len(nums1) + len(nums2) + 1) / 2  mid
. This guarantees that the elements in the left half are less than or equal to the elements in the right half.d. Median Calculation: If the max element on the left of both partitions is less than or equal to the min element on the right of both partitions, calculate the median. If the total number of elements is even, it’s the average of the max element on the left and the min element on the right. If it’s odd, it’s the max element on the left.
e. Binary Search Adjustment: If we haven’t found the correct partition, adjust the binary search bounds based on the comparison of max elements on the left and min elements on the right. If max of left elements in nums1 is greater than min of right elements in nums2, move towards left in nums1 (adjust max_index). Else, move towards right (adjust min_index).
Independent Parts: The binary search on the two arrays is the main part of this problem and it’s not divisible further. However, the median calculation based on whether the total length is even or odd can be considered as an independent step after the correct partition has been found.
Repeatable Patterns: The process of adjusting the binary search bounds based on comparison of elements near the partition is a repeatable pattern until the correct partition is found. The entire approach is an application of the Binary Search pattern on two sorted arrays.
Solution Approach and Analysis
Let’s use an analogy to understand this problem better. Imagine you are a captain of a large ship sailing in the ocean at night, and you’re trying to find the middle point between two lighthouses which are at different distances from your ship, both providing sorted light beams, similar to our sorted arrays.
Step 1  Prepare for the Journey: You first check the size of the light beams (arrays). If one of the lighthouses (arrays) is not emitting any light (empty), you know the middle point is simply the middle of the other lighthouse’s light beam. If not, you ensure to start your journey towards the lighthouse that has shorter light beam (smaller array). This is because it will take less time (lower complexity) to navigate.
Step 2  Set Sail: Now, you’re sailing towards the shorter lighthouse, you reach midway, and also calculate a corresponding point in the other lighthouse’s beam such that total distance covered in both light beams is equally divided.
Step 3  Look Around: Once you reach the mid point, you compare the light beam intensity (numbers in the array) around you. If the last bit of light beam behind you from both lighthouses is less intense than the first bit of light beam in front of you, then you know you’ve found the middle point. If the total length of the combined light beam is even, the middle point is average of the last bit of light beam behind you and the first bit of light beam in front of you. If it’s odd, the middle point is just the last bit of light beam behind you.
Step 4  Adjust Course: If you have not found the middle point in Step 3, you need to adjust your position. If the last bit of light beam behind you from the shorter lighthouse is more intense than the first bit of light beam in front of you from the longer lighthouse, you need to move back in the shorter lighthouse’s light beam. If the last bit of light beam behind you from the longer lighthouse is more intense, you need to move forward in the shorter lighthouse’s light beam.
Step 5  Repeat: Repeat Step 3 and Step 4 until you find the middle point.
Let’s demonstrate this with an example. Consider the arrays nums1 = [1, 3, 8] and nums2 = [2, 7, 12]. They’re both sorted.
We start in the middle of nums1 (since it’s the shorter array), so our initial partition is around the number 3. In nums2, we partition around the number 7, since the total length of the left side of both partitions should be equal to the total length of the right side.
We look at the elements near the partition: maxLeft1 = 3, minRight1 = 8, maxLeft2 = 2, and minRight2 = 7.
We find that maxLeft1 > minRight2. This means that we’re too far to the right in nums1, so we adjust our binary search to move to the left. We repeat this process until we find that maxLeft1 <= minRight2 and maxLeft2 <= minRight1.
Once we find the correct partition, we calculate the median: since the total number of elements is odd, it’s max(maxLeft1, maxLeft2) = max(3, 2) = 3.
So, the median of the two arrays is 3.
Thought Process
Given the constraints and requirements of the problem, we have to design an efficient approach that leverages the sorted nature of the input arrays. The problem’s mention of a logarithmic time complexity suggests we should think about binary search or other divideandconquer strategies.
Here is a step by step thought process:
Step 1  Problem Understanding: We are tasked with finding the median of two sorted arrays. The median of a sorted list is the middle element if the length is odd, or the average of the two middle elements if the length is even.
Step 2  Leverage Sorted Arrays: We can take advantage of the fact that the arrays are sorted. This property is particularly beneficial for binary search.
Step 3  Use Binary Search: Due to the requirement of logarithmic time complexity, binary search is an excellent candidate. The binary search should be applied to the shorter array of the two to keep the time complexity to a minimum.
Step 4  Divide and Conquer: The key is to partition both arrays in such a way that the maximum of the left partition is less than or equal to the minimum of the right partition, and both partitions have equal numbers of elements (or differ by just one when the total length is odd).
Step 5  Calculate Median: Depending on whether the total length of elements in the two arrays is odd or even, calculate the median from the elements in the partition.
Here’s a Python solution that follows the thought process:


This solution ensures that the time complexity is O(log(min(m, n))) as it always performs the binary search on the smaller array. It works by partitioning the two arrays such that all the elements in the left partition are less than all the elements in the right partition. The median is then found based on the elements at the partition boundaries.
From Brute Force to Optimal Solution
Let’s begin with a brute force solution and gradually optimize it.
Brute Force Solution:
The most straightforward approach to find the median of two sorted arrays would be to merge the two arrays into a new one, sort it, and then simply pick the median. In Python, this would look something like:


In this brute force solution, the inefficiency arises from merging and sorting the two arrays. The time complexity of sorting is O((m + n) log(m + n)), where m and n are the lengths of the two input arrays. For large m and n, this can be quite slow.
Optimized Solution:
The key insight for optimization is realizing that you don’t need to sort both entire arrays to find the median. Since the input arrays are already sorted, we can utilize a binary search to find the median.
We apply the binary search on the shorter array, calculate a partition
such that elements on the left side of the partition in both arrays are less than the elements on the right side. We adjust the partition
until we find the correct position. The median is then calculated based on the elements around the partition
. The time complexity is reduced to O(log(min(m, n))) because the binary search is applied only on the shorter array.
Here’s the optimized Python code for the problem:


By applying this binary search method, we’ve optimized the time complexity of the solution from O((m + n) log(m + n)) to O(log(min(m, n))). This is a significant improvement for large m and n, and is the best we can do for this problem. The space complexity remains O(1), as we are not using any extra space that scales with input size.
Code Explanation and Design Decisions
Initial parameters: The initial parameters in this case are the two sorted arrays
nums1
andnums2
. We are looking for the median element(s) across both these arrays, so they form the entirety of our solution domain. The variablex
is the length ofnums1
andy
is the length ofnums2
.Primary loop: The primary loop here is a binary search loop running on the smaller of the two input arrays. Each iteration of this loop represents an attempt to find a suitable partition of the two arrays such that all elements on the left are less than all elements on the right. If such a partition is found, it means that we have located the correct position of the median, fulfilling the requirement of the problem.
Conditions or branches: There are a few conditions inside the loop, and they help to adjust the position of the partition based on comparison of elements on both sides of the partition. The conditions ensure that the median position maintains its requirement of having all elements on the left smaller than the ones on the right, across both arrays.
Updates or modifications: Within the loop, the start or end of the binary search range gets updated based on the outcome of the partition comparison. If
maxLeftX
is more thanminRightY
, it means the partition is too far to the right, so we move the end to the left. If not, it means the partition is too far to the left, so we move the start to the right. These modifications reflect the adjustments made to locate the correct position of the median.Invariant: The invariant here is the sorted nature of the two arrays. Despite the changes to the partition position, the arrays
nums1
andnums2
remain sorted throughout the execution. This invariant is crucial for the binary search and partition strategy to work correctly.Final output: The final output is the median of the two sorted arrays. Depending on the total length of the combined array (whether it’s odd or even), the median could be the average of two middle elements or it could be the middle element itself. This result is the solution to our problem as it fulfills the requirement of finding the median of two sorted arrays.
Coding Constructs
Highlevel strategies: The main strategy used by this code is binary search. It’s applied in the context of finding an optimal partition point in the two arrays such that all elements on the left side of the partition are less than all elements on the right. The sorted nature of the input arrays makes binary search an efficient strategy for this problem.
Explaining to a nonprogrammer: This code is like a librarian looking for a specific page in two sorted books. Instead of going page by page, the librarian starts at the middle, and based on whether the desired page is earlier or later, the search continues in the respective half of the book. The goal here is to find a page in each book such that all pages before these are earlier than all pages after them.
Logical constructs: This code uses a loop for repeated actions (the binary search), conditions to change behavior based on the data (checking and adjusting partition position), and mathematical operations to calculate indices and median values.
Algorithmic approach in plain English: The code first identifies the smaller of the two arrays to minimize the search space. Then it initiates a binary search where it calculates the partition point in both arrays such that the elements on the left are always less than the elements on the right. If such a point is found, the median is calculated based on these partitions. If the point isn’t found, the search space is reduced and the process is repeated.
Key steps or operations: The key steps include deciding the start and end points of the binary search, calculating partition points in the two arrays, comparing elements around the partition to check if the correct partition point is found, and adjusting the search space based on the comparison results.
Algorithmic patterns: The key algorithmic pattern here is the binary search. It uses a divide and conquer strategy to efficiently find a target value in a sorted array, or in this case, an optimal partition point in two sorted arrays.
Language Agnostic Coding Drills
Array Manipulation  Difficulty: Low. This is the basic understanding of how to manipulate and use arrays, such as accessing elements, understanding indices, and calculating lengths.
Mathematical Operations  Difficulty: Low. This involves basic arithmetic calculations like addition, division and use of floor operation.
Variable Assignment and Update  Difficulty: Low. This includes understanding how to assign and update variable values in a program, such as assigning initial search boundaries and updating them in each iteration.
Conditional Statements  Difficulty: Low to Medium. Understanding how to use ifelse conditions to control the flow of the code based on certain conditions.
Loop Constructs  Difficulty: Medium. The use of a while loop to keep repeating a sequence of steps. The loop continues until a certain condition is met.
Binary Search  Difficulty: High. This is the heart of the solution, the binary search algorithm. It involves understanding the logic of dividing the search space in half in each iteration based on some condition.
The problemsolving approach starts with understanding the problem and the properties of the input  two sorted arrays. From the problem statement, we deduce that we need to find a partition point in the arrays such that elements on one side are all smaller than elements on the other.
The concept of a ‘partition’ is derived from the requirement to find the median  a value that separates a distribution into two equal halves. But instead of merging the arrays and then finding the partition, we seek an efficient approach due to the sorted nature of the arrays and the time complexity constraint.
This leads us to the Binary Search drill. The key insight is that we can apply binary search not just to find a specific target value, but to find a suitable partition point. We start the binary search on the smaller array for efficiency. The partition points in both arrays are adjusted such that the highest element on the left is smaller than the lowest element on the right.
The drills of using conditional statements and loop constructs come into play to implement the binary search, with the mathematical operations and variable assignment/update drills being used inside the loop. Finally, once a suitable partition is found, the median is calculated using mathematical operations based on whether the combined length of arrays is even or odd.
This is how the individual drills fit together to form the final solution, showing how smaller coding concepts can be used to solve a complex problem.
Targeted Drills in Python
 Variable and Function Definitions:


 Conditional Statements:


 Loops:


 Lists and Indexing:


 Float and Integer Division:


 Arithmetic Operations and Comparisons:


 Classes and Method Definitions:


Let’s create Pythonbased drills for each of the coding concepts identified earlier.
 Array Manipulation This is the drill for basic array manipulation. We’ll create an array and perform common operations like accessing elements and calculating the length.


 Mathematical Operations Here we will do simple arithmetic operations like addition and division, and the use of floor operation.


 Variable Assignment and Update Assigning initial values to variables and updating them.


 Conditional Statements Using ifelse conditions to control the code flow.


 Loop Constructs Use of while loop to repeat a sequence of steps.


 Binary Search Implementing a basic binary search to find a target number in a sorted array.


After mastering these individual drills, the next step is to integrate them in a logical order to solve the problem. We need to:
 Use array manipulation to handle inputs and set initial partition points for binary search.
 Use binary search, controlled by the while loop, to find the partition point where all elements on the left are smaller than elements on the right. This is done by continually updating variables based on conditional statements.
 Once the partitions are found, calculate the median using mathematical operations based on whether the total length of arrays is even or odd.
Assembling these drills in the correct order and manner leads to the final solution of the problem.
Q&A
Technical correctness is the best type of correctness.
a sample illustrates that [1,2,3,4] ought to yield 2.5. However, the STATISTICAL DEFINITION of median is the DATA POINT which seperates the upper half from the lower half. By definition, the median MUST be a member of the data set.
Bad question.
The definition of the median does indeed involve splitting the dataset so that half the numbers are lower and half the numbers are higher. However, in a dataset with an even number of values, there isn’t a single data point that separates the upper half from the lower half. In such a case, statisticians and mathematicians typically take the mean (average) of the two middle numbers.
For example, in the array [1, 2, 3, 4], the numbers 2 and 3 are the two middle numbers. There isn’t a single number in the array that separates the upper half (3, 4) from the lower half (1, 2). So, the median is calculated as the mean of 2 and 3, which is 2.5.
So while the median is typically a member of the dataset in cases where the number of data points is odd, it is not necessarily so when the number of data points is even. This problem requires the implementation of a solution that handles both cases correctly.
The question states: “There are two sorted arrays nums1 and nums2 of size m and n respectively.” This does not tell me whether the sort order is largest to smallest, or smallest to largest. It would even be possible for nums1 to be sorted largest to smallest while nums2 is sorted from smallest to largest. So I thought we would need to be able to handle such cases, making no assumptions as to the direction of the sorts. As a test case of opposite sort directions in the inputs I tried: [10,8,3] [5,9,17]
Both input lists sorted as required, but in opposite directions. The median here should be 8.5, but when I ran the test case it gave me an expected answer of 6.5. So I think leetcode is making an assumption about the sort directions of the inputs that should not be made, and it is not designed to properly handle this case. What are we allowed to assume about the input sort direction?
The problem statement does not explicitly mention the sort order. However, it is conventionally assumed that, unless stated otherwise, a “sorted” array or list is in nondecreasing order, meaning from smallest to largest. This assumption is consistent with most problems and discussions involving sorted data in computer science and mathematics.
Therefore, in the context of this problem, you can assume that both input arrays are sorted in nondecreasing order. This assumption informs the approach to solving the problem. If the arrays were sorted in nonincreasing order or if one array was sorted in increasing order while the other in decreasing order, a different approach might be necessary.
However, if the problem were to handle arrays sorted in either direction, a preprocessing step could be added to sort the arrays in the desired order before proceeding with the rest of the algorithm. This would increase the time complexity, though, because sorting is generally an O(n log n) operation.
It would have been great of you would have given clear examples for median of 2 sorted arrays.
Some examples for clarity:
Example 1:
Input: nums1 = [1, 3] nums2 = [2]
The merged array would be [1, 2, 3], and the median is 2.
Example 2:
Input: nums1 = [1, 2] nums2 = [3, 4]
The merged array would be [1, 2, 3, 4], and since it’s an evensized array, we take the average of the two middle numbers. So, the median is (2 + 3) / 2 = 2.5.
Example 3:
Input: nums1 = [1, 5, 9] nums2 = [2, 6, 10]
The merged array would be [1, 2, 5, 6, 9, 10]. Since there are 6 elements, we take the average of the 3rd and 4th elements, which are 5 and 6. So, the median is (5 + 6) / 2 = 5.5.
Example 4:
Input: nums1 = [1, 7, 9] nums2 = [10, 11]
The merged array would be [1, 7, 9, 10, 11]. As there are 5 elements, the median is the middle element, which is 9.
These examples show how to find the median for two sorted arrays. They represent different scenarios, including even and oddsized merged arrays, and they demonstrate how to calculate the median in each scenario.
Why the test case with nums1=[3,4] and nums2=[] have median = 3.00000 whereas it should be 3.50000?
Given the arrays nums1 = [3, 4] and nums2 = [], the median would actually be 3.5, not 3.0.
The definition of median is the middle number in a sorted, ascending or descending, list of numbers. If the list contains an even number of observations, then there is no distinct middle value and the median is then defined to be the arithmetic mean of the two middle values. In this case, the two middle values are 3 and 4. So, the median is (3 + 4) / 2 = 3.5.
If you’re seeing a different result, there may be an error in the code or algorithm that’s being used to calculate the median.
Why do we need to swap nums1 and nums2 if sizeof(nums1) < sizeof(nums2). If I don’t do that it gives runtime error for the case [2] [] but I am not getting why.
If the size of nums1 is smaller than the size of nums2, we swap them so that the binary search always happens on the smaller array (nums1). It is an optimization step that reduces the time complexity of the algorithm by reducing the search space.
When dealing with the case nums1 = [2] and nums2 = [], if you didn’t perform the swap, you might be running into issues because your binary search or indexing might be trying to access elements of an empty array (nums2 in this case). This will cause a runtime error because you’re trying to access a location that doesn’t exist.
By ensuring nums1 is the smallest array, you ensure that binary search is always performed on a nonempty array, which avoids runtime errors.
Remember, the core principle here is to conduct the binary search on the smaller array. This not only optimizes the algorithm but also prevents potential errors when one of the arrays is empty.
Any help would be appreciated if you could explain why the solution in the O(log(min(m,n))) case requires a swap if the second array is smaller.
If you could specify any edge case that is failing , that would be of great help.
Swapping nums1 and nums2 if nums1 is larger than nums2 is a trick to simplify the problem by reducing it to a condition where we always binary search on the smaller array. This doesn’t change the result because the median of two arrays doesn’t depend on which array is longer, but it makes the code cleaner and easier to understand, and it also improves performance by minimizing the search space.
Moreover, this technique ensures that we do not run into an index out of bounds error during our binary search process.
Let’s consider an edge case where the swapping would be important:
nums1 = [6, 7, 8, 9, 10] nums2 = [1, 2, 3]
If we did not swap, and started the binary search on nums1, we would end up exploring unnecessary parts of nums1 that wouldn’t impact the final result, since we know that the median must be within the range covered by nums2 due to its smaller size and range of values.
If we swap and binary search on nums2, we can find the median more efficiently because we focus on the smaller range of values. We then only need to compare with a few specific elements of nums1.
To summarize, the swap is used to ensure efficiency and simplicity by always performing binary search on the smaller array.
Can someone explain to me how this is the answer?
Input [3] [2,1] Output 0.75 Expected 1.0
My output is .75. the median of nums1 is 3, the median of nums2 is 1.5. Therefore (median 1 + mediant 2 / 2) = .75. How do they get 1?
Your confusion seems to stem from a misunderstanding of the definition of the median for an aggregated set, rather than individual medians.
In this question, we’re looking for the median of the combined sorted array, not the average of the medians of the individual arrays. This is a key difference.
Let’s see how it works with your example:
 nums1 = [3]
 nums2 = [2, 1]
When you combine these arrays, you get:
combined = [2, 1, 3]
The median of this combined array is the middle element, which is 1, not 0.75.
So, the expected output of 1 is correct because 1 is the median of the combined sorted array, not the average of the individual medians. The median is defined as the value separating the higher half from the lower half of a data sample. It does not mean averaging medians of individual sets.
Look up definition of median. You want to find the value that bigger than half and smaller than half. In the 1 sorted array, this is the value at index ~n/2, but we have 2 sorted arrays case here. My first idea is to look at median of the 1st array, and median of the second array, you can easily infer that the median of both must bigger than the min of them and smaller than the max of them, hence you can eliminate half of each array (i.e. left half of the array with smaller median). Now we are looking for a O(log(n+m)) solution, we want to eliminate half (or some fixed percentage) each iteration. So the goal is try to eliminate as many as possible while keeping track of what have been eliminated to balance the bigger half and the smaller half, I reached a dead end with this process. What works for me:
1st hint: from above, I came to realize to solve this, I should approach a more generalize problem: instead of finding the median, find the xth position value (basically bigger than x other values and smaller than (n+mx1) other values). if that’s solved, then the median will just be x=(n+m)/2; in case n+m is not even, use floor and ceiling, and then average. 2nd hint: no need trying to eliminate from both side, I only try to eliminate the left side. My recursive solution reduces x each time, till it reaches x = 0 (basic case) 3rd hint: inspired by the 1st elimination attempt, my solution is: pick the index i1 for 1st array and index i2 for 2nd array, in a clever way, such that when comparing these 2 values, you can eliminate 1st i1 values of 1st array or the 1st i2 values of 2nd array.
The hints you provided give a clear pathway towards the solution. I can expand on them further.
This hint is a key step towards solving the problem. When we are looking for the median of a sorted list, we typically look for the element at index n/2 for an odd length list, and average of elements at index n/2 and n/2  1 for an even length list. However, in the case of two sorted lists, we want to generalize this by finding the kth smallest element where k is (n+m)/2 for odd length combined list and [(n+m)/2] and [(n+m)/2  1] for an even length combined list. The problem is thus transformed into finding the kth smallest element in two sorted arrays.
Your second hint is an optimization of the general solution. Instead of trying to eliminate elements from both arrays, you focus on eliminating elements from one side. This is easier to manage and reason about, and doesn’t compromise on the solution’s efficiency.
The third hint is about the method of elimination. Given indices i1 and i2 in the first and second arrays respectively, if the element at i1 is smaller, then all elements at indices less than or equal to i1 in the first array cannot be the kth smallest element, as there are already i1 elements in the first array and i2 elements in the second array that are smaller. Thus, they can be safely eliminated. The same reasoning applies if the element at i2 is smaller. This elimination process significantly reduces the search space, and brings us closer to the solution.
Given these hints, the solution can be implemented using a binary search algorithm. The key is to eliminate elements that we are certain cannot be the kth smallest element, thereby gradually reducing our search space until we find the desired element. The complexity of the solution is O(log(min(m,n))), since in each iteration we eliminate half of the elements from one of the arrays.
This problem nicely demonstrates how a seemingly complex problem can be transformed into a simpler one, and efficiently solved using a standard algorithm with some key modifications.
Similar Problems
Here are 10 problems that use similar underlying concepts:
Search in Rotated Sorted Array  This problem requires binary search on a rotated sorted array, very similar to finding the partition points in our original problem.
First Bad Version  You need to apply binary search to minimize the number of calls to an expensive API.
Find Peak Element  This problem also requires binary search on an array. It also involves finding a partition point like our original problem.
Find Minimum in Rotated Sorted Array  The solution involves binary search on a rotated sorted array.
Search for a Range  This problem involves doing a binary search twice on a sorted array to find the start and end of a range.
Sqrt(x)  This problem requires you to implement int(sqrt(x)) and one of the efficient ways is to use binary search.
Find K Closest Elements  This problem uses binary search to find the correct position in the array, similar to our original problem.
Smallest Good Base  This problem uses binary search to minimize the base, given a number and length of representation.
Minimum Size Subarray Sum  The solution uses binary search for the subarray size, similar to how we used it for finding the partition point.
Koko Eating Bananas  This problem applies binary search in a range of possible solutions.
These require applying the binary search algorithm, which was a key aspect in the solution of the original problem.