Top K Frequent Elements
Key Insights:
Frequency Count: You need to find the frequency of each unique integer in the array.
Return K Elements: You need to return the ‘k’ most frequent integers.
Uniqueness Guaranteed: The problem ensures that the answer will be unique, so there are no ties for the ‘k’ most frequent elements.
Range and Limits: Pay attention to constraints. The array length can be up to 105, and the elements can range from 104 to 104.
Output Order: The output doesn’t need to be sorted. Any order is acceptable.
K Range: ‘k’ will always be valid; it’s between 1 and the total number of unique elements.
Multiple Steps: The problem can be broken into multiple steps like counting frequencies, sorting based on frequency, and then selecting the top ‘k’ elements.
To solve this problem, you can approach it in parts: count frequencies, sort them, and then select the top ‘k’ elements.
Visualizing the problem can help break it down into simpler components. Here are some ways to visualize this problem:
Frequency Table: Imagine a table where one column represents the unique integers in the array, and the other column represents their frequency.
 For
[1, 1, 1, 2, 2, 3]
, the table would look like:Number  Frequency  1  3 2  2 3  1
 For
Bar Chart: Create a bar chart where the xaxis represents the unique integers and the yaxis represents their frequencies.
3  # 2  # # 1  # # #  1 2 3
Priority Queue: If you plan to use a data structure like a heap to solve the problem, you can visualize it as a “bucket” that holds the elements sorted by frequency.
k Elements: Imagine a separate container where you will collect the ‘k’ most frequent elements. It starts empty and fills up as you find these elements.
Steps: Write down or visualize the sequence of steps you need to perform. This can help clarify the process.
 Count frequencies
 Sort by frequency
 Pick top ‘k’
By visualizing in this manner, you can see each part of the problem more clearly, which helps in solving it efficiently.
In an abstract representation, this problem can be described as follows:
Given a sequence ( S ) consisting of ( n ) elements ( s_1, s_2, …, s_n ) and an integer ( k ), find a set ( R ) containing ( k ) elements that appear most frequently in ( S ).
Key Elements:
 ( S )  The input sequence.
 ( n )  The length of the sequence ( S ).
 ( k )  The number of elements to be returned in the set ( R ).
 ( R )  The output set containing ( k ) elements with the highest frequency.
Constraints:
 ( 1 \leq n \leq 10^5 )
 ( 10^4 \leq s_i \leq 10^4 )
 ( k ) is between ( 1 ) and the number of unique elements in ( S ).
Operations:
 Frequency Counting: Calculate the frequency of each unique element in ( S ).
 Sorting: Sort the unique elements based on their frequency.
 Selection: Select the top ( k ) elements based on sorted frequency.
Objective:
To find the set ( R ) satisfying the above conditions.
The abstract representation encapsulates the core logic and constraints of the problem, stripped of any realworld specifics.
Frequency Count: This term refers to the number of times each unique element appears in the input array. In this problem, frequency counting is the first step in identifying the ‘k’ most frequent elements.
Sorting: Ordering elements based on certain criteria. Here, sorting is done based on the frequency of elements in the array.
Integer Array: An ordered collection of integers. In this problem, the integer array
nums
serves as the input data set.Constraint: A rule or limit that defines the conditions that the input or output must meet. Understanding the constraints (like the range for ‘k’ and array length) is essential for crafting an efficient solution.
Priority Queue (Heap): A specialized treebased data structure that keeps its elements in sorted order. It can be used to efficiently find the ‘k’ most frequent elements after counting frequencies.
Unique Element: An element that appears only once in a set. In the context of this problem, ‘k’ is limited to the number of unique elements in the array.
Set: A collection of unique elements. The output can be thought of as a set of ‘k’ most frequent integers, although the problem allows for the output to be in any order.
BigO Notation: A way to describe the performance of an algorithm. Knowing the constraints allows you to choose an algorithm that meets the performance requirements (like O(n log n) or O(n)).
Each of these terms plays a role in defining the problem space, specifying what operations need to be performed, or how to optimize the solution.
Let’s break this down.
Key Concepts:
 Frequency Count: Count how many times each number shows up in the list.
 Sorting: Put these numbers in order, from the ones that show up most often to the ones that show up least.
 Top ‘k’: Pick the first ‘k’ numbers from this sorted list.
How They Interact:
 First, you count how often each number appears.
 Next, you sort these numbers by how often they show up.
 Finally, you pick the first ‘k’ numbers from your sorted list.
Metaphor:
Imagine you’re a teacher, and you have a box of different kinds of fruit stickers to give to your students. Each kind of sticker represents a unique number. Every time you give a sticker to a student, you make a tally mark next to that kind of sticker on a sheet of paper. At the end of the day, you want to know which ‘k’ types of stickers are the most popular among your students, based on how many tally marks each type has.
 The tally sheet is your “Frequency Count,” where you keep track of each sticker type’s popularity.
 Then you rearrange this list of sticker types so that the most popular ones are at the top and the least popular ones are at the bottom. This is “Sorting.”
 Finally, you look at the top ‘k’ sticker types on your sorted list to find out which stickers are the most popular. This is picking the “Top ‘k’.”
In this analogy, the stickers are the numbers in your list, the tally marks are the frequency count, and the top ‘k’ sticker types are the ‘k’ most frequent numbers you’re looking for.
Specific Characteristics for Efficient Solution:
Frequency Count Exploitation: Since you need to find the frequency of elements, a single pass through the array to count occurrences can be highly efficient (O(n)).
Uniqueness Constraint: The answer is guaranteed to be unique. This removes the need for handling tiebreaking scenarios, making the algorithm simpler.
Range of ‘k’: The value of ‘k’ will always be between 1 and the number of unique elements. This range restriction can make the process of finding the ‘k’ most frequent elements faster as ‘k’ will never be more than the unique elements in the array.
Element Range: Elements in the array are between 10^4 and 10^4. Knowing this range could allow for arraybased counting methods if appropriate, although the range is quite large.
Size Limitations: The maximum array length is 10^5, which informs us that algorithms of complexity O(n log n) or better are likely to be suitable. Quadratic or worse algorithms may not be efficient enough.
Output Order: The output elements can be in any order, eliminating the need for an additional sorting step for the final output.
Integer Values: All values are integers, which are easier and faster to compare and sort than floatingpoint numbers.
Lower Limit: The lower constraint of 1 for ’nums.length’ and ‘k’ means we don’t have to account for empty arrays or zero values for ‘k’, simplifying edge cases.
These characteristics can guide you in selecting the appropriate data structures and algorithms, such as hash tables for frequency counts and heaps or quickselect for finding the ‘k’ most frequent elements efficiently.
Key Insights from Constraints:
Upper Bound of Array Length: The maximum array length is (10^5). This suggests that an algorithm with time complexity of (O(n \log n)) or better should be suitable for solving the problem efficiently.
Element Range: The elements range from (10^4) to (10^4). While this range is quite large, it informs us that we are working with integers, which are easier to handle than floatingpoint numbers.
Lower Bound: The minimum value for both ’nums.length’ and ‘k’ is 1, which simplifies the problem by removing the need to handle empty arrays or a zero value for ‘k’.
‘k’ Range: ‘k’ will be between 1 and the number of unique elements in the array. This constraint helps us to eliminate any potential errors or inefficiencies related to invalid ‘k’ values.
Uniqueness Guarantee: The uniqueness of the answer means we don’t have to consider tiebreaking scenarios. This can simplify the implementation and improve efficiency.
Any Order for Output: The output can be in any order, freeing us from the need to sort the final ‘k’ elements, which can be a timesaver.
Array Element Limit: The lower and upper bounds for the array elements are given, which may help in selecting the most appropriate data structure for counting frequencies.
Guaranteed Answer: The problem guarantees that an answer will exist within the constraints, so there’s no need to check for “no solution” scenarios.
Understanding these constraints allows us to optimize the solution by selecting appropriate data structures and algorithms, ensuring they fall within the bounds of the problem’s requirements.
Let’s explore various examples and test cases that capture different aspects of the problem space.
Categorized Cases:
1. Minimum Input Case (Edge Case)
 Input: nums = [1], k = 1
 Output: [1]
 Analysis: The smallest possible array size and ‘k’ value. In this case, the array contains only one element, and ‘k’ is also 1. This tests if the algorithm can handle minimum inputs.
2. All Elements Unique (Edge Case)
 Input: nums = [1, 2, 3, 4, 5], k = 2
 Output: [1, 2] or any other combination of two numbers
 Analysis: All elements are unique, so any two numbers will be the ‘most frequent’ (each appears once). This tests the algorithm’s ability to handle unique elements and the ‘any order’ output requirement.
3. All Elements Same (Edge Case)
 Input: nums = [1, 1, 1, 1, 1], k = 1
 Output: [1]
 Analysis: All elements are the same. This tests whether the algorithm can handle scenarios where there’s only one unique number.
4. Multiple Frequencies (General Case)
 Input: nums = [1, 2, 2, 3, 3, 3], k = 2
 Output: [2, 3]
 Analysis: The elements have different frequencies, and ‘k’ is less than the number of unique elements. This is a more general case and tests the core functionality.
5. Negative Numbers (General Case)
 Input: nums = [1, 1, 2, 2, 1], k = 2
 Output: [1, 1] or [1, 2] or [1, 2]
 Analysis: This includes negative numbers and tests if the algorithm can handle both positive and negative integers correctly.
6. ‘k’ Equals Number of Unique Elements (Boundary Case)
 Input: nums = [1, 1, 2, 3], k = 3
 Output: [1, 2, 3]
 Analysis: In this case, ‘k’ is equal to the number of unique elements. This will test if the algorithm can handle such boundary conditions.
7. Multiple Top Frequencies, Smaller ‘k’ (Edge Case)
 Input: nums = [1, 1, 2, 2, 3, 3], k = 1
 Output: [1] or [2] or [3]
 Analysis: Here, multiple numbers have the same top frequency, but ‘k’ is smaller than the number of these top elements. This tests how the algorithm handles tie scenarios when ‘k’ is smaller.
Analyzing these test cases should help ensure a robust solution that can handle a variety of scenarios, conforming to the problem constraints and requirements.
Visualizing these cases can help to understand them better. Here’s how to think about each one visually:
1. Minimum Input Case (Edge Case)
 Visualization:
[1]
 ‘k’:
[1]
Think of a single bucket containing the number ‘1’. ‘k’ asks for one bucket, and you only have one, so you take it.
2. All Elements Unique (Edge Case)
 Visualization:
[1, 2, 3, 4, 5]
 ‘k’:
[1, 2]
Imagine having five separate buckets, each containing one of the numbers. ‘k’ asks for any two buckets, so you can choose any.
3. All Elements Same (Edge Case)
 Visualization:
[1, 1, 1, 1, 1]
 ‘k’:
[1]
Picture a single, overflowing bucket with the number ‘1’. You only need one bucket for ‘k’, and you have just one to take.
4. Multiple Frequencies (General Case)
 Visualization:
[1, 2, 2, 3, 3, 3]
 ‘k’:
[2, 3]
Visualize this as three buckets: one with one ‘1’, another with two ‘2’s, and a third with three ‘3’s. ‘k’ asks for the two fullest buckets.
5. Negative Numbers (General Case)
 Visualization:
[1, 1, 2, 2, 1]
 ‘k’:
[1, 1]
Imagine four buckets: two have one ‘1’, one has one ‘1’, and one has one ‘2’. ‘k’ asks for any two of these buckets, but the bucket with ‘1’ has the most stickers.
6. ‘k’ Equals Number of Unique Elements (Boundary Case)
 Visualization:
[1, 1, 2, 3]
 ‘k’:
[1, 2, 3]
You have three buckets: one with two ‘1’s, another with one ‘2’, and a third with one ‘3’. ‘k’ asks for all the buckets, so you take all.
7. Multiple Top Frequencies, Smaller ‘k’ (Edge Case)
 Visualization:
[1, 1, 2, 2, 3, 3]
 ‘k’:
[1]
Imagine three equally filled buckets: one with two ‘1’s, another with two ‘2’s, and a third with two ‘3’s. ‘k’ asks for just one of these buckets, so you can choose any.
The visualization helps you to think about the array as ‘buckets of frequencies’, making it easier to conceptualize what ‘k’ most frequent elements mean.
Key Insights from Analyzing Different Cases:
Handling Minimum Inputs: The algorithm should be able to deal with the smallest possible inputs, like a single element and ‘k’ equal to 1.
Unique Elements: When all elements are unique, the algorithm should be flexible enough to return any ‘k’ elements since they all have the same frequency.
Uniform Frequency: In cases where all elements have the same frequency, the algorithm needs to be adept at selecting any ‘k’ elements.
Multiple Frequencies: The core functionality of finding ‘k’ most frequent elements is tested here. The algorithm should correctly identify elements with higher frequencies.
Negative Numbers: The presence of negative numbers should not affect the algorithm’s ability to find the most frequent elements. It should be equally competent in handling both positive and negative integers.
‘k’ Equals Number of Unique Elements: The algorithm should be capable of returning all unique elements when ‘k’ equals the number of unique elements in the array.
Multiple Top Frequencies, Smaller ‘k’: The algorithm should be able to handle situations where multiple elements share the highest frequency but ‘k’ is smaller than the number of these top elements.
Output Order Irrelevant: In all cases, the algorithm should remember that the order of output elements doesn’t matter, potentially saving computation time.
Unique Output: The answer will be unique, so no need to consider tiebreaking criteria, which simplifies the algorithm.
By understanding these insights, you can ensure that your solution is robust and handles all edge and boundary conditions effectively.
Asking clarification questions can help to fully understand the problem and any hidden nuances. Here are some questions that could be considered:
Output Ordering: Is it necessary for the elements in the output to be sorted in any particular way, such as by frequency or value, even though the problem states it can be in any order?
Frequency Ties: In case multiple elements have the same frequency, is there a preference for which ones to choose? Or is any selection with the correct frequency acceptable?
Input Constraints: Are there any additional constraints on the input array? For example, could it contain zero, or is it strictly positive and negative integers?
Memory Constraints: Are there any limitations on the amount of additional memory that can be used?
Mutable Input: Is it acceptable to modify the input array, or does it need to remain unchanged?
Time Complexity: Is there a target time complexity to aim for, beyond the general guideline provided by the size of the input array?
Return vs Output: Does the problem expect the solution to return the ‘k’ most frequent elements, or is printing them sufficient?
Empty Array: The problem specifies that the array length is at least 1, but what should the behavior be if this constraint is violated? Should the code handle it or assume it won’t happen?
‘k’ Value: The problem specifies that ‘k’ will be at least 1 and less than or equal to the number of unique elements. Is it safe to assume that ‘k’ will always be a valid integer within these constraints?
Understanding these points can help in crafting a solution that not only solves the problem but also handles various edge conditions and constraints effectively.
A variety of mathematical and algorithmic concepts can be leveraged to tackle this problem effectively:
Hashing: Using a hash table to count the frequency of each element in the array makes this task efficient. Hashing provides constanttime access and insertion operations.
Priority Queue: After counting the frequencies, you can use a minheap priority queue to keep track of the top ‘k’ most frequent elements. This allows you to maintain a ‘ksized’ heap, making it more efficient than sorting all elements.
Sorting: Sorting by frequency after counting them could also be an approach, but it’s generally less efficient (O(n log n)) than using a heap (O(n log k)).
Bucket Sort: If the frequency range is known and small, you can use bucket sort to achieve linear time complexity.
Divide and Conquer: This paradigm isn’t directly applicable here but could be used for variations of the problem where sorting might be essential.
Order Statistics: The problem relates to finding ‘k’ largest elements (by frequency) from a set. This is similar to the kth order statistics problem, often solved using QuickSelect algorithm.
Counting Sort: If the range of the elements is restricted, counting sort could be an efficient way to sort elements. However, for this problem, it may be less useful due to the wide range of integer values (10^4 to 10^4).
TimeSpace Tradeoff: Utilizing extra space for a hash table results in a faster algorithm. This is a classic example of a timespace tradeoff.
Greedy Algorithms: While not directly applicable, the nature of selecting the ‘k’ most frequent elements lends itself conceptually to greedy methods.
By applying one or more of these mathematical or algorithmic concepts, you can make the problem more manageable and likely find an efficient solution.
Let’s say you have a basket full of different fruits like apples, oranges, and bananas. Some fruits appear more often in the basket than others. For example, you might have 5 apples, 3 oranges, and 2 bananas.
Now, you want to know which fruits show up the most. If someone asks you for the top 2 most frequent fruits, you would pick apples and oranges, because you have more of them compared to bananas.
In this problem, instead of fruits, you have a list of numbers. Some numbers appear more than once. You need to find out which numbers show up the most times, according to how many ’top numbers’ someone asks for. So if they ask for the ’top 2’, you give them the 2 numbers that appear the most in the list.
Just like how you can count each type of fruit in your basket to know which is the most common, you count each number in the list to find out which numbers appear the most. Then, you pick the top ones based on what is asked.
That’s the idea of this problem: to find out which numbers in a list show up the most times.
Certainly, let’s break down how to find the k most frequent elements in an integer array stepbystep.
Steps:
Count Frequency:
 The first step is to count how many times each number appears in the list.
 Think of this like tallying how many apples, oranges, and bananas you have in a fruit basket.
Store FrequencyElement Pairs:
 After counting, keep track of each number and its frequency.
 Imagine writing down on a piece of paper: “Apple: 5, Orange: 3, Banana: 2”.
Prioritize Top Frequencies:
 Now, sort the numbers based on how often they appear, but you only need the top ‘k’.
 Think of making a “Top Fruits” chart but only with room for ‘k’ fruits.
Extract Topk Elements:
 Finally, take the top ‘k’ frequent numbers from the sorted list.
 In our metaphor, pick up the top ‘k’ fruits and put them on the chart.
How to Do It:
Hash Table for Counting:
 Use a hash table to count the frequency of each number.
frequency_count = {1: 3, 2: 2, 3: 1}
for an input like[1, 1, 1, 2, 2, 3]
.
MinHeap for Topk:
 Use a minheap (priority queue) to keep track of the top ‘k’ frequent elements.
 Heap elements would look like
(frequency, element)
.
Populate Heap:
 Go through the frequency hash table. For each element:
 If the heap size is less than ‘k’, insert it.
 If the heap size is ‘k’, compare the smallest (root) with the current element. If current is greater, remove root and insert the current element.
 This ensures only top ‘k’ remain.
 Go through the frequency hash table. For each element:
Collect Output:
 The heap now contains the top ‘k’ frequent elements. Extract them.
Effect of Changes in Parameters:
 Larger ‘k’: More elements to keep in the heap, which could slow down the process.
 Larger Array: More elements to count initially, taking up more time and space.
 More Unique Elements: Increases the time complexity because you have to look at more elements.
Example Case:
 Input:
[1, 1, 1, 2, 2, 3]
, k = 2  Step 1: Count frequency
{1: 3, 2: 2, 3: 1}
 Step 2: Initialize minheap
 Step 3: Populate heap
 Insert
(3, 1)
(3 is the frequency of 1)  Insert
(2, 2)
(2 is the frequency of 2)  Don’t insert
(1, 3)
as it’s smaller
 Insert
 Step 4: Extract
(3, 1)
and(2, 2)
from heap  Output:
[1, 2]
By following these steps, you can effectively find the ‘k’ most frequent elements in any integer array.
the key terms and their roles in solving this problem are:
Integer Array (
nums
): The list of numbers you’re given.
 This is your main input and suggests you’ll need to traverse it at least once to count frequencies.
Integer (
k
): Specifies how many of the most frequent elements to find.
 Tells you the size constraint for your top ‘k’ elements, guiding you towards using a minheap of size ‘k’ for efficiency.
Frequency:
 How often each number appears in the list.
 Directs you to use a hash table for counting frequencies for quick lookups.
Most Frequent Elements:
 The main output you’re interested in.
 This is what you’ll extract from your minheap at the end of the algorithm.
Hash Table:
 A data structure used for storing keyvalue pairs.
 Useful for counting the frequency of elements in constant time.
MinHeap (Priority Queue):
 A binary tree for which every parent node has a value less than or equal to any of its children.
 Helpful for keeping track of the top ‘k’ most frequent elements without sorting all elements.
Any Order (Output):
 The problem states that the output elements can be in any order.
 This relieves the need for sorting the output, making the algorithm more efficient.
Unique Answer:
 The problem guarantees that the answer is unique.
 This simplifies the problem by ensuring that you don’t have to consider multiple valid outputs.
Constraints:
 Numeric limits on the input size and elements.
 These guide you on the expected efficiency of your solution and how much memory it should use.
Each term guides specific choices: using a hash table for counting frequencies and a minheap for maintaining the top ‘k’ elements, avoiding unnecessary sorting, and ensuring the solution meets the input constraints for size and time complexity.
Visual aids like tables and diagrams can help you identify and understand the key properties of this problem. Here’s how you can use them:
Frequency Table:
 To recognize the frequency of each element in the integer array (
nums
), create a table with two columns: one for the unique elements and another for their frequencies.  This table is a visual representation of the hash table used in the algorithm.
Element  Frequency  1  3 2  2 3  1
 To recognize the frequency of each element in the integer array (
Heap Diagram:
 Draw a tree structure to represent the minheap.
 This can help you understand how the minheap prioritizes elements based on frequency.
(2, 2) / (3, 1)
Workflow Diagram:
 Create a flowchart to understand the steps of the algorithm.
 Boxes can represent operations like ‘Count Frequency’, ‘Initialize Heap’, ‘Populate Heap’, etc., and arrows can show the flow of data between them.
[Start] > [Count Frequency] > [Initialize MinHeap] > [Populate Heap] > [Collect Output] > [End]
Constraint Table:
 Use a table to list all the constraints and their implications on the solution.
Constraint  Implication  1 <= nums.length <= 105  Solution must handle large datasets 10^4 <= nums[i] <= 10^4  Numbers can be negative or positive
KValue Consideration:
 You can draw a number line or even a bar graph to visualize different ‘k’ values and their respective top ‘k’ most frequent elements.
k=1: [1] k=2: [1, 2]
Each of these visual tools helps you to recognize different properties or steps in solving the problem, making it easier to develop and debug your algorithm.
1. Stepwise Refinement:
Initialize Data Structures: 1.1. Create an empty hash table
frequency_count
. 1.2. Initialize an empty minheapmin_heap
.Count Frequencies: 2.1. Loop through each element in
nums
. 2.2. Updatefrequency_count
for each element.Populate MinHeap: 3.1. Loop through
frequency_count
. 3.2. Ifmin_heap
size < k, insert the element. 3.3. Else, compare and possibly replace the smallest element inmin_heap
.Extract Solution: 4.1. Create an empty list
output
. 4.2. Emptymin_heap
intooutput
.
2. Granular, Actionable Steps:
Initialize Data Structures:
frequency_count = {}
min_heap = []
Count Frequencies:
1 2 3 4 5
for num in nums: if num in frequency_count: frequency_count[num] += 1 else: frequency_count[num] = 1
Populate MinHeap:
1 2 3 4 5 6 7
for num, freq in frequency_count.items(): if len(min_heap) < k: heapq.heappush(min_heap, (freq, num)) else: if freq > min_heap[0][0]: heapq.heappop(min_heap) heapq.heappush(min_heap, (freq, num))
Extract Solution:
1 2 3
output = [] while min_heap: output.append(heapq.heappop(min_heap)[1])
3. Independent Parts:
Frequency Counting:
 This step can be done independently of the rest, as it only depends on the input array
nums
.
 This step can be done independently of the rest, as it only depends on the input array
Heap Population:
 This part needs the
frequency_count
but is otherwise independent of the extraction step.
 This part needs the
4. Repeatable Patterns:
Element Traversal:
 We repeatedly loop through elements, first in
nums
and then infrequency_count
.
 We repeatedly loop through elements, first in
Heap Operations:
heapq.heappush()
andheapq.heappop()
are repeatedly used to managemin_heap
.
Conditional Checks:
 Checks on the size of
min_heap
and the frequency of elements are performed in a pattern within the loop.
 Checks on the size of
By breaking down each part, you can focus on solving independent chunks of the problem, making it easier to understand, code, and debug.
Approach to Solving the Problem
Step 1: Initialize Data Structures
Start by initializing two key data structures:
 A hash table called
frequency_count
to store each element’s frequency.  A minheap called
min_heap
to store the top ‘k’ most frequent elements.
Analogy:
Think of frequency_count
as a tally sheet where you mark how many times each number shows up. min_heap
is like a leaderboard where only the top ‘k’ performers are displayed.
Step 2: Count Frequencies
 Loop through each element in the input array
nums
.  Update its frequency in
frequency_count
.
This is the first critical step because without knowing the frequency of each element, we can’t find the ‘k’ most frequent ones.
Step 3: Populate MinHeap
 Loop through the elements in
frequency_count
.  For each element:
 If the heap size is less than ‘k’, add it to
min_heap
.  If the heap is full, compare the current frequency with the smallest frequency in the heap. If it’s larger, remove the smallest and insert the new one.
Analogy:
Imagine min_heap
as a contest with ‘k’ podium positions. As new contestants (elements) try to get on the podium, they have to beat the lowest performer to earn a spot.
Step 4: Extract Solution
 Initialize an empty list
output
.  Extract elements from
min_heap
intooutput
.
How Parameters Affect the Solution
 Array Length: A longer array will increase the time to count frequencies but won’t necessarily impact the minheap operations, as it’s constrained by ‘k’.
 Range of Array Elements: The range (10^4, 10^4) doesn’t directly affect the algorithm’s efficiency but defines the range of possible keys in
frequency_count
.  Value of ‘k’: A larger ‘k’ will require more space for the minheap and more time for heap operations.
Example Case:
Let’s take the example where nums = [1,1,1,2,2,3]
and k = 2
.
Initialize:
frequency_count = {}
min_heap = []
Count Frequencies:
frequency_count = {1: 3, 2: 2, 3: 1}
Populate MinHeap:
 Insert (3, 1), heap becomes [(1, 3)]
 Insert (2, 2), heap becomes [(1, 3), (2, 2)]
 Insert (3, 1), heap becomes [(1, 3), (2, 2), (3, 1)]
 Extract Solution:
output = [3, 1]
So, the output is [3, 1]
, which are the ‘k=2’ most frequent elements in the array. Note that the order does not matter.
An invariant in this problem is the size of the min_heap
, which should not exceed ‘k’ after the heap is populated initially. Throughout the algorithm, the heap is maintained to store the top ‘k’ most frequent elements based on their frequency.
This invariant guides the logic of the heap operations, ensuring that we are on track to find the correct ‘k’ most frequent elements. After inserting an element into a full heap of size ‘k’, one element (the one with the smallest frequency) is removed to maintain this invariant. This way, we ensure that min_heap
always contains the ‘k’ most frequent elements seen so far.
Basic Thought Process and Steps
Step 1: Understand the Problem
The first cue in the problem statement is “return the k most frequent elements.” This immediately tells us that we need to count the frequency of each element in the array. It also suggests sorting or some method to find the “most frequent” elements.
Insight:
Finding frequency points towards using a hash table for fast lookup. Sorting all elements based on frequency would be inefficient, so consider a data structure that can maintain the top ‘k’ elements, like a heap.
Step 2: Plan the Approach
 Use a hash table to store the frequency of each element.
 Use a minheap to store the ‘k’ most frequent elements.
Step 3: Code the Solution
Here is a Python code based on the above approach.


This code accomplishes the task of finding the ‘k’ most frequent elements from the given list nums
. It leverages a hash table for quick frequency lookup and a minheap to efficiently maintain the ‘k’ most frequent elements.




This code defines a Solution
class with a method topKFrequent
that takes an integer list nums
and an integer k
. It returns a list of ‘k’ most frequent elements in nums
. The method uses a hash table for frequency counting and a minheap for keeping track of the ‘k’ most frequent elements.
1. Parameters:
 Inputs: The method takes two parameters:
nums
andk
.  Types:
nums
: List of integers (List[int]
)k
: Integer (int
)
 Context:
nums
represents the list of numbers where we want to find the ‘k’ most frequent elements.k
specifies how many of the most frequent elements we want.
2. Preconditions:
 State of the Program: No specific state of the program is required before calling this method.
 Constraints:
 1 <= nums.length <= 10^5
 10^4 <= nums[i] <= 10^4
 k is in the range [1, the number of unique elements in
nums
].
3. Method Functionality:
 Expected Behavior: The method is expected to return a list of ‘k’ most frequent integers from the input list
nums
.  Interaction: The method reads the input list
nums
and integerk
and processes them to return the result.
4. Postconditions:
 State of the Program: No change in the state of the program or input parameters.
 Return Value: Returns a list of integers representing the ‘k’ most frequent elements in
nums
.  Side Effects: None.
5. Error Handling:
 Preconditions Not Met: Since the problem specifies constraints, the function assumes that the inputs meet these conditions. If they do not, the behavior is undefined.
 Exception Handling: The method doesn’t throw any exception or return a special value as it assumes that the inputs meet the constraints mentioned in the problem.
1. Problem Understanding:
The problem asks for the ‘k’ most frequent elements from a given list of integers. You need to count the occurrences of each integer in the list and then return the ‘k’ integers that appear most frequently.
2. Initial Breakdown:
The major parts or stages are:
 Counting the frequency of each integer in the list.
 Identifying the ‘k’ most frequent integers.
3. Subproblem Refinement:
Counting Frequency:
 Iterate through the list.
 Use a hash table to store the count of each integer.
Identifying ‘k’ Most Frequent:
 Use a minheap to keep track of ‘k’ most frequent elements.
 Populate the heap and then extract the elements.
4. Task Identification:
 Iterate and count: This is a general task applicable to many problems.
 Heap operations: Minheap insertion and extraction are generalized tasks.
5. Task Abstraction:
 Iterate and Count: Loop through a list and count the occurrences of each element. This task is abstract enough and reusable.
 Heap Operations: Insert and extract elements from a minheap based on frequency. This is also abstract and reusable.
6. Method Naming:
countFrequencies
: For counting the frequency of each integer.findTopKElements
: For identifying the ‘k’ most frequent integers using a minheap.
7. Subproblem Interactions:
 Order: Start with
countFrequencies
and then proceed tofindTopKElements
.  Dependencies:
findTopKElements
depends on the frequencies counted bycountFrequencies
.
By identifying these subproblems and tasks, we can formulate an approach to solving the problem efficiently.
Brute Force Solution
Step 1: Count Frequencies
 Use a hash table to count the frequency of each number in the list
nums
.  Time Complexity: (O(N))
 Use a hash table to count the frequency of each number in the list
Step 2: Sort Based on Frequency
 Sort the hash table entries by their frequency.
 Time Complexity: (O(N \log N))
Step 3: Extract ‘k’ Most Frequent
 Extract the first ‘k’ elements from the sorted list.
 Time Complexity: (O(k))
Brute Force Python Code:


Inefficiencies:
 Sorting all frequencies is unnecessary. We only need the top ‘k’.
 Time complexity: (O(N \log N))
 Space complexity: (O(N))
Optimized Solution
Step 1: Count Frequencies
 Same as before ((O(N)) time).
Step 2: Use a MinHeap for Top ‘k’
 Use a minheap to keep track of the ‘k’ most frequent elements.
 Populate the heap with the first ‘k’ elements.
 For each subsequent element, compare it with the root of the minheap, which is the smallest among the top ‘k’. Replace the root if the new element has higher frequency.
 Time Complexity: (O(N \log k))
Optimized Python Code:


Reasoning:
 Using a minheap to track the top ‘k’ elements avoids sorting all elements, thereby reducing time complexity.
Time and Space Complexity:
 Time Complexity: (O(N \log k))
 Space Complexity: (O(N)) (hash table), (O(k)) (minheap)
The optimized solution reduces the time complexity from (O(N \log N)) to (O(N \log k)) while maintaining the same space complexity of (O(N)).
1. Initial Parameters
nums
: A list of integers. Represents the data set we need to analyze to find the most frequent elements.k
: An integer. Specifies how many top frequent elements we’re interested in.
2. Primary Loop or Iteration
In both the bruteforce and optimized solutions, the primary iteration is over the list nums
to create a frequency count (hash table). Each iteration counts the occurrence of an integer from nums
.
BruteForce
 Iterates through the sorted list of keyvalue pairs (frequency, number) to extract the top ‘k’ most frequent numbers.
Optimized
 Iterates over the minheap to maintain the top ‘k’ frequent numbers.
3. Conditions or Branches
No specific conditions or branches within the primary loop in both solutions. The frequency counting and heap operations handle all elements uniformly.
4. Updates or Modifications to Parameters
 The frequency hash table gets updated with each iteration over
nums
, reflecting changes in the state of the solution.  In the optimized solution, the minheap may get updated (root replaced) as we discover numbers with higher frequencies.
5. Invariant
 In both solutions, the hash table maintains the accurate frequency count of each unique number in the list
nums
.  In the optimized solution, the minheap always contains ‘k’ elements, maintaining the smallest frequency at the root.
6. Final Output
The final output is a list of ‘k’ integers. These are the ‘k’ most frequent numbers from the list nums
, fulfilling the problem’s requirements. In the optimized solution, the use of a minheap ensures efficiency, delivering the solution in (O(N \log k)) time.
1. HighLevel Strategies
The code uses two main techniques:
 Frequency counting to determine how many times each unique element appears in the input array.
 Heap data structure to maintain the ‘k’ most frequent elements efficiently.
2. Purpose for a NonProgrammer
The code finds out which numbers show up most often in a list and returns the top ‘k’ frequent ones.
3. Logical Elements
 Iteration: To loop through the list of numbers.
 Hash Table: To store and track the frequency of each number.
 MinHeap: To keep track of the top ‘k’ frequent numbers.
 Conditionals: To decide whether to update the minheap or not.
4. Algorithmic Approach in Plain English
 Count how often each number appears in the list.
 Put the ‘k’ most frequent numbers in a special container (minheap).
 Go through the list of frequencies.
 For each frequency, if it’s greater than the smallest in the special container, swap them.
 At the end, the special container has the ‘k’ most frequent numbers.
5. Key Steps or Operations
 Loop through
nums
to populate the frequency hash table.  Initialize a minheap with the first ‘k’ frequent numbers.
 Loop through the remaining frequencies to maintain the minheap with top ‘k’ frequent numbers.
6. Algorithmic Patterns
 Frequency Counting: A common pattern for problems requiring the counting of elements.
 Heap Management: Used for efficiently tracking top ‘k’ elements, a variant of the Selection algorithm.
1. Distinct Coding Concepts
Variable Initialization: The most basic concept, setting up variables to hold values.
Iteration: Looping over arrays or lists to perform repeated actions.
Conditional Statements: Making decisions based on conditions.
Data Structure: Array/List: Storing multiple values in an organized way for quick access.
Hash Table/Dictionary: Storing keyvalue pairs for quick lookup.
Heap Operations: Understanding and manipulating heap data structures.
Frequency Counting: Using a hash table to keep track of the count of elements in an array.
Heap Management for ‘k’ Elements: Using a heap to efficiently manage the ‘k’ most frequent elements.
2. Order of Increasing Difficulty and Descriptions
Variable Initialization: Easiest and most fundamental concept.
Data Structure: Array/List: Basic concept but introduces multiple elements and indexing.
Iteration: Requires understanding loops, but not too complex as you’re usually iterating over known structures like arrays.
Conditional Statements: Adds logical branching, making the code more complex to trace.
Hash Table/Dictionary: A step up in complexity, requiring an understanding of keyvalue pairs and hash functions.
Frequency Counting: Built on top of Hash Table and Iteration, this requires a higher level of understanding.
Heap Operations: Introduces a new set of rules and behaviors for data manipulation. It requires a good understanding of data structures.
Heap Management for ‘k’ Elements: The most complex, as it combines heap operations with other tasks like frequency counting.
3. ProblemSolving Approach
Variable Initialization: Start by setting up necessary variables. These could be counters, accumulators, or storage for temporary values.
Data Structure: Array/List: Create an array to hold elements you’ll read from the input.
Iteration: Loop through the array to inspect each element. This loop will help us count the frequency of each element.
Hash Table/Dictionary: While iterating through the list, populate a hash table with frequency counts of each element in the list.
Conditional Statements: Use conditions to decide whether to add a new element in the hash table or update the frequency of an existing element.
Frequency Counting: Finish the first pass through the array to complete the frequency count in the hash table.
Heap Operations: Initialize a minheap with the ‘k’ most frequent elements from the hash table.
Heap Management for ‘k’ Elements: Loop through the remaining frequencies, and use heap operations to maintain the heap with the ‘k’ most frequent elements.
By combining these drills, you can incrementally build up the logic to solve the problem, starting from basic array manipulation and ending with complex heap management operations. Each drill adds a new layer of complexity to the overall solution.
1. Pythonbased Coding Drills for General Concepts
Variable Initialization


Data Structure: Array/List


Iteration


Conditional Statements


Hash Table/Dictionary


Frequency Counting


Heap Operations


Heap Management for ‘k’ Elements


2. ProblemSpecific Coding Drills
Retrieve ‘k’ Most Frequent Elements
It’s essential for our problem because we need to find the ‘k’ most frequent elements.


3. Integration of Drills to Solve the Initial Problem
Variable Initialization: Initialize a hash table to hold frequency counts and a heap to hold the ‘k’ most frequent elements.
Data Structure: Array/List: Read the array data, which holds the numbers whose frequencies we need to find.
Iteration + Hash Table/Dictionary: Loop through the array, populating the hash table with the frequency of each number.
Conditional Statements + Frequency Counting: Use conditional statements to increment the frequency count in the hash table while iterating.
Heap Operations: Create a heap from the frequencies stored in the hash table.
Heap Management for ‘k’ Elements: Apply heap operations to maintain the heap containing the ‘k’ most frequent elements.
Retrieve ‘k’ Most Frequent Elements: Use heap operations to retrieve the ‘k’ most frequent elements, which is our final output.
After you have implemented each drill separately, integrating them should follow the steps as described, leading you to the final solution. Each drill serves as a building block to your final code.
Identifying Problem Isomorphism
“Top K Frequent Elements” falls under a category of hash maps for frequency counting and priority queues for maintaining the top K elements.
A simpler problem in this category is “Single Number”. This problem simply requires you to find a single element that doesn’t appear twice in an array. Frequency counting can be used to solve this problem, making it a simpler variant.
A more complex problem in this category is “LFU Cache”. This problem not only requires frequency counting, but it also needs a way to track the least frequently used element in a cache. This necessitates a more complex combination of data structures, increasing the difficulty level.
The reasoning behind these choices:
 “Single Number” is simpler than “Top K Frequent Elements” because it involves only frequency counting and not maintaining a set of top K elements.
 “LFU Cache” is more complex than “Top K Frequent Elements” because it combines frequency counting with cache eviction policies, which requires a more sophisticated usage of data structures.
Therefore, the order of problems from simpler to more complex, based on frequency counting and usage of advanced data structures, would be:
 “Single Number”
 “Top K Frequent Elements”
 “LFU Cache”
This mapping is an approximation. While these problems share themes around frequency counting and data structure usage, the exact problem specifics and constraints can significantly impact the complexity and the approaches to solution.
Problem Classification
This problem can be classified as follows:
Data Analysis: This problem involves analysing the frequency of integers in an array, which falls under the category of data analysis.
Array Manipulation: As the data structure you’re working with is an array, this problem also involves array manipulation.
Sorting/Ranking: You’re being asked to return the k most frequent elements, which is a form of sorting or ranking problem. You’re not sorting the entire array, but you are identifying the top k elements according to some criteria.
Counting: The problem involves counting the frequency of each element in the array.
Selection: The task requires you to select a subset of elements (the k most frequent ones) from a larger set (the array).
Parameterized Complexity: The problem complexity depends on an external parameter (k), apart from the size of the input (n).
The actual solution will likely involve hash tables or priority queues, and the complexity constraint suggests a need for an algorithm more efficient than a naive sorting approach.
Steps
To solve this problem, you could take the following steps:
Understand the Problem: The first step in solving any problem is understanding it. In this case, you’re given an array of integers and a value k. The task is to find the k most frequent integers in the array and return them in any order.
Devise a Plan: Once you’ve understood the problem, you need to figure out how to solve it. One approach to solve this problem is by using a frequency counter and a bucket sort.
Code: Here’s how you can implement the solution in Python:


This approach first counts the frequencies of each number, then sorts the numbers into “buckets” based on their frequencies. Finally, it gathers the most frequent elements from these buckets. Since the buckets are iterated in reverse order (i.e., from highest frequency to lowest), this ensures that the most frequent elements are gathered first.
Testing: After coding, you should always test your code with multiple test cases to ensure it’s working as expected. This is also the time to debug and fix any issues that may arise.
Optimize: If the code is working as expected and meets the problem constraints, you’re done! Otherwise, you may need to optimize your solution.
Remember that the problem requires a solution that’s faster than O(n log n). Since the solution here involves a single pass to count frequencies and another pass to gather the most frequent elements, it’s time complexity is O(n), which meets the problem’s requirements.
Concepts in Increasing Level of Difficulty
To solve this problem, several programming concepts and techniques are needed. Here they are, listed in increasing order of difficulty:
Variables and Data Types: Understanding how to use variables and different data types (in this case, integers and lists) is fundamental to any programming task.
Loops: You need to know how to use loops to iterate over arrays/lists. Both “for” and “while” loops are useful in this problem.
Conditional Statements: Using “if” statements is essential to check certain conditions during execution.
Functions: Understanding how to define and use functions is crucial in structuring your code and implementing the solution.
Arrays/Lists: You need to understand how to work with arrays or lists, as the input is provided in this format. This includes creating, indexing, and manipulating them.
Counting with Dictionaries or Hashmaps: This problem involves counting the frequencies of elements. This is often done using a dictionary or hashmap, where the keys are the elements to count, and the values are their counts.
Sorting: Understanding sorting algorithms is useful in many problems. In this case, it’s not the main solution approach, but sorting related knowledge can help in understanding the bucket sort concept.
Bucket Sort: This problem is solved efficiently using a variant of the bucket sort algorithm. Understanding how bucket sort works is probably the most complex part of this problem.
Complexity Analysis: Lastly, being able to analyze time and space complexity is important in assessing the efficiency of your solution. This problem requires a solution faster than O(n log n).
Alternative Solution


Language Agnostic Coding Drills
To break down the solution into its smallest units for learning, the following concepts and drills could be isolated and implemented separately:
Variable Initialization: Understand how to declare and initialize variables in your chosen programming language.
Drill: Write a program that declares and initializes a few variables of different types and then prints them.
Lists and Arrays: Learn how to create lists or arrays, and understand how to index, append, and manipulate them.
Drill: Write a program that creates an array, adds elements to it, removes elements from it, and accesses elements by index.
Loops: Learn about different types of loops (for, while, etc.) and when to use each. Understand how to iterate over arrays or lists with a loop.
Drill: Write a program that uses loops to print out the elements of an array.
Creating a Function/Method: Understand the syntax of defining a function or method, how to call it, and what return values are.
Drill: Write a function that accepts some inputs, performs a calculation or manipulation on them, and returns a result.
Dictionary and Counter: Learn about dictionary data structure (keyvalue pairs) and how to create, manipulate, and access them. Also, understand how Counter works in counting the frequency of elements.
Drill: Create a dictionary, add keyvalue pairs to it, access the value by key, and remove keyvalue pairs. Use Counter to count frequency of elements in a list.
Nested Loops: Understand how nested loops work and when to use them.
Drill: Write a program with nested loops, such as printing a 2D array.
Conditionals: Learn how to use ifelse statements to control the flow of your program.
Drill: Write a program that makes decisions based on the values of variables using ifelse statements.
List Comprehensions: Understand how to use list comprehensions for creating new lists.
Drill: Create a list using list comprehension by performing some operation on each element of an existing list.
Sorting and Reversing Lists: Understand how to sort and reverse lists or arrays.
Drill: Write a program that sorts and reverses a list or array.
Combining Concepts: Combine all the learned concepts to solve a complex problem.
Drill: Reassemble all the drills to form the final solution, i.e., create a method that uses a counter, list comprehension, nested loops, conditionals, and return statement to find the k most frequent elements in an array.
This approach allows learning to be broken down into manageable chunks, with each drill focusing on one aspect of the overall problem. This way, the learning process becomes more structured and manageable, allowing learners to gradually build up their understanding and skillset.
Solution for Coding Drills in Python
Here is simple Python implementations of each of the drills:
Variable Initialization
1 2 3
a = 5 b = "Hello, world!" print(a, b)
Lists and Arrays
1 2 3 4
my_list = [1, 2, 3] my_list.append(4) my_list.pop(0) print(my_list[1])
Loops
1 2 3
numbers = [1, 2, 3, 4, 5] for num in numbers: print(num)
Creating a Function/Method
1 2 3 4
def add_numbers(a, b): return a + b print(add_numbers(3, 4))
Dictionary and Counter
1 2 3 4 5 6 7 8 9 10
from collections import Counter my_dict = {"apple": 1, "banana": 2} my_dict["orange"] = 3 print(my_dict["banana"]) del my_dict["apple"] my_list = ["apple", "apple", "banana", "banana", "banana"] cnt = Counter(my_list) print(cnt)
Nested Loops
1 2 3 4
matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] for row in matrix: for num in row: print(num)
Conditionals
1 2 3 4 5 6 7 8
a = 5 b = 10 if a < b: print("a is less than b") elif a > b: print("a is greater than b") else: print("a is equal to b")
List Comprehensions
1 2 3
original_list = [1, 2, 3, 4, 5] squared_list = [num ** 2 for num in original_list] print(squared_list)
Sorting and Reversing Lists
1 2 3 4 5
my_list = [3, 1, 4, 1, 5, 9] my_list.sort() print(my_list) my_list.reverse() print(my_list)
Combining Concepts
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
class Solution: @staticmethod def topKFrequent(nums: List[int], k: int) > List[int]: cnt = Counter(nums) buckets = [[] for _ in range(len(nums) + 1)] for val, freq in cnt.items(): buckets[freq].append(val) res = [] for bucket in reversed(buckets): for val in bucket: res.append(val) k =1 if k == 0: return res
Each of these drills focuses on one specific concept, helping to build understanding and familiarity before moving onto the next one. By the end, the learner should be able to combine the concepts to solve the original problem.
Here are 10 problems that use similar underlying concepts:
Top K Frequent Words
 This problem is closely related because it also involves finding the top ‘k’ elements based on frequency, but in this case, the elements are words instead of numbers.
Find All Anagrams in a String
 Involves using a frequency count, similar to the original problem, to find all anagrams of a substring in a larger string.
First Unique Character in a String
 It requires frequency counting to identify characters that appear only once.
Intersection of Two Arrays II
 Utilizes hash tables to keep track of the frequency of elements in arrays, similar to how frequency is used in the original problem.
Sort Characters By Frequency
 This problem also uses frequency counting but adds an extra step where the elements need to be sorted based on their frequency.
Kth Largest Element in an Array
 This problem is related due to its usage of heaps to find the ‘kth’ largest element, similar to finding the top ‘k’ frequent elements.
Find K Pairs with Smallest Sums
 It uses heap data structures to manage ‘k’ smallest elements, similar to our original problem that involves heap management for ‘k’ elements.
Contains Duplicate II
 Utilizes a hash table to track element frequencies along with their indices to satisfy specific constraints, somewhat similar to frequency counting in the original problem.
Sliding Window Maximum
 Uses a heap to track the maximum elements in a sliding window across an array, requiring heap operations similar to our original problem.
Kth Smallest Element in a Sorted Matrix
 Similar to the original problem, this problem also involves using heap data structures to extract ‘kth’ elements but in a 2D matrix.
Each of these problems involves either frequency counting, heap management, or hash tables, which are key aspects of the original problem.
Q&A
Is O(n * k) better than O(n log n)?
The comparison between O(n * k) and O(n log n) depends on the values of ( n ) and ( k ):
If ( k ) is small, or close to a constant, O(n * k) can be similar to or better than O(n log n). For example, if ( k = 2 ), then O(n * k) = O(2n) = O(n), which is better than O(n log n).
If ( k ) is larger than ( log(n) ), then O(n * k) will be worse than O(n log n). For example, if ( k = n ), then O(n * k) = O(n^2), which is worse than O(n log n).
If ( k = log(n) ), both complexities would be equivalent: O(n * log(n)) = O(n log n).
So, whether O(n * k) is better than O(n log n) is contextdependent.
Here is a table comparing O(n * k) and O(n log n) for different values of ( n ) and ( k ).
( n )  ( k )  ( n \log n )  ( n \times k )  Comment on O(n * k) vs O(n log n) 

10  2  23  20  O(n * k) is better 
100  2  460  200  O(n * k) is better 
1000  2  6907  2000  O(n * k) is better 
10  10  23  100  O(n * k) is worse 
100  50  460  5000  O(n * k) is worse 
1000  100  6907  100000  O(n * k) is worse 
1000  7  6907  7000  O(n * k) is better 
1000  10  6907  10000  O(n * k) is worse 
This table clearly shows that the efficiency of O(n * k) in comparison to O(n log n) varies depending on the values of ( n ) and ( k ). When ( k ) is small, O(n * k) tends to be better, and when ( k ) is large, it tends to be worse.