Find First and Last Position of Element in Sorted Array
To find the starting and ending position of a given target value in a sorted array with (O(\log n)) runtime complexity, you can use binary search twice. First, you find the starting position of the target value, and then you find the ending position.
Here’s the code:


 The
binary_search
function takes three parameters:left
,right
, andfind_first
. Iffind_first
isTrue
, it finds the starting position of the target; otherwise, it finds the ending position.  If the target is found, it either looks to the left or right to find the first or last occurrence.
 If the target is not found, the function returns
[1, 1]
.
This solution meets the constraint of (O(\log n)) runtime complexity.
Identifying Problem Isomorphism
“Find First and Last Position of Element in Sorted Array” can be mapped to “Search for a Range”.
Both require the application of binary search in a sorted array to find the starting and ending position of a given target value. The search must be done twice: once for the starting position of the target, and once for the ending position.
In “Find First and Last Position of Element in Sorted Array”, you are given a sorted array of integers and a target value, and your task is to find the starting and ending position of the target value in the array.
In “Search for a Range”, you are given a sorted array of integers and a target value, and you need to find the range that the target spans in the array. If the target does not exist in the array, you should return [1, 1].
Both have the same level of complexity and are the same, asking for the range of a target value in a sorted array. The tasks are so similar that they can be seen as different wordings of the same problem.


Problem Classification
This is a search problem in the domain of computer science algorithms and data structures.
The ‘What’ components are:
 An ordered integer array as input
 A target integer value to search for
 Finding the start and end indices of the target in the array
 Must have O(log n) runtime
Based on this, I would categorize it as:
Domain: Search algorithms
Problem type: Range search query on ordered integer data
Subtype: Binary search candidate
Explanation:
The ordered array structure allows leveraging binary search.
We need to find a target range efficiently in logarithmic time.
Binary search provides a way to rapidly hone in on a range in ordered data.
So in summary, this is a range search problem over ordered data, falling under the domain of search algorithms. The time constraint and ordered array imply that binary search is likely the intended algorithm category.
Language Agnostic Coding Drills
Defining Classes and Functions: The ability to define classes and methods is fundamental to objectoriented programming. In most languages, you define a class using a keyword like
class
and define methods within that class. The functionsearchRange
is a method defined inside theSolution
class.Working with Arrays (or Lists): Understanding how to create and manipulate arrays or lists is essential in most modern programming languages. Here,
nums
is a list of integers and you access its elements and length.Arithmetic operations: Division (
//
) and addition (+
) operations are used here to calculate the mid index of the list.Variable Assignment: This code involves basic variable assignments, which are operations that assign a value to a variable. Here,
lo
,hi
, andmid
are variables that get assigned different values at different points.Loops: This code includes a while loop, which is a control flow statement that allows code to be executed repeatedly based on a given Boolean condition. The while loop inside the
search
function continues untillo
is no longer less thanhi
.Conditional Statements: This code uses ifelse statements, which are used to perform different actions based on different conditions. Here, they are used to check the condition of the elements in
nums
array and adjust thelo
andhi
pointers.Comparisons: The code uses comparisons (
<
,<=
) to decide which code branches to execute.Returning from Functions: The code includes the use of the
return
statement, which ends the function execution and optionally returns a value.
Arranging these concepts from simpler to more complex, we can derive the following order:
 Variable Assignment
 Arithmetic Operations
 Defining Classes and Functions
 Working with Arrays (or Lists)
 Comparisons
 Conditional Statements
 Loops
 Returning from Functions
Targeted Drills in Python
Variable Assignment
Write a Python code to assign values 10 and 20 to variables
a
andb
respectively.1 2 3
a = 10 b = 20 print(a, b)
Arithmetic Operations
Create a Python function that takes two arguments and returns their division result.
1 2 3 4
def divide(a, b): return a // b print(divide(20, 5)) # output: 4
Defining Classes and Functions
Define a Python class
Calculator
with a methodadd
that takes two numbers and returns their sum.1 2 3 4 5 6
class Calculator: def add(self, a, b): return a + b calc = Calculator() print(calc.add(3, 4)) # output: 7
Working with Arrays (or Lists)
Write a Python code to create a list of integers and print the length of the list.
1 2
nums = [1, 2, 3, 4, 5] print(len(nums)) # output: 5
Comparisons
Create a Python function that compares two numbers and returns the larger one.
1 2 3 4 5 6 7
def max_num(a, b): if a > b: return a else: return b print(max_num(3, 7)) # output: 7
Conditional Statements
Write a Python function that checks if a number is even or odd.
1 2 3 4 5 6 7 8
def check_even_odd(num): if num % 2 == 0: return 'Even' else: return 'Odd' print(check_even_odd(4)) # output: Even print(check_even_odd(7)) # output: Odd
Loops
Create a Python function that prints all the numbers from 1 to n.
1 2 3 4 5 6 7
def print_numbers(n): i = 1 while i <= n: print(i) i += 1 print_numbers(5)
Returning from Functions
Write a Python function that takes a list of integers and returns their sum.
1 2 3 4 5 6 7
def sum_of_list(nums): total = 0 for num in nums: total += num return total print(sum_of_list([1, 2, 3, 4, 5])) # output: 15
All of these drills can be run individually and later combined for a more complex code implementation, like in the provided code snippet.
10 Prerequisite LeetCode Problems
The “34. Find First and Last Position of Element in Sorted Array” problem involves performing a binary search in a sorted array to find the start and end index of a particular element. It requires a good understanding of binary search and array manipulation.
Here are 10 simpler problems to prepare:
278. First Bad Version: This problem involves performing a binary search to find the first bad version in a sorted list of versions.
35. Search Insert Position: This problem asks you to determine the index where a target value should be inserted in a sorted array.
69. Sqrt(x): In this problem, you perform binary search to compute and return the square root of a specified number.
744. Find Smallest Letter Greater Than Target: This problem involves searching for the smallest letter in a sorted list that is greater than a given target letter.
704. Binary Search: This problem is a direct application of binary search in a sorted array.
441. Arranging Coins: This problem can be solved by applying binary search concept where you are to find the total number of full staircase rows that can be formed with a given number of coins.
167. Two Sum II  Input array is sorted: In this problem, you are to find two numbers such that they add up to a specific target number.
349. Intersection of Two Arrays: This problem involves finding the intersection of two arrays. Though it doesn’t directly use binary search, it still helps with understanding array manipulations.
367. Valid Perfect Square: This problem involves using binary search to determine if a number is a perfect square.
374. Guess Number Higher or Lower: This problem involves using binary search to guess a number within a given range based on feedback.
Problem Analysis and Key Insights
Here are some key insights gained from analyzing the problem statement:
The array is sorted in nondecreasing order  this property can be exploited to optimize search using binary search.
We want to find the start and end indices for a target value  so the problem is a range search rather than just lookup.
There may be multiple instances of the target value  so we need to find the boundaries.
The problem constraints require O(log n) runtime  this points to binary search as the intended algorithm.
The array size can be up to 100,000 elements  so an efficient algorithm is needed to meet the log n constraint.
The target may not exist in the array  the [1, 1] output signifies invalid range.
The integer constraints allow pruning optimizations during comparison.
The key insights are recognizing this is a range search problem over ordered data for which binary search is wellsuited, along with constraints that underscore the need for an efficient logarithmic time algorithm.
Problem Boundary
Based on the problem statement, here is how I would summarize the scope:
Inputs: Ordered integer array, target integer value
Output: Start and end indices of target in array, [1,1] if not found
Objective: Find indices bounding occurrences of target in the sorted array
Assumptions:
 Array is sorted in nondecreasing order
 Array contains no duplicates
 Target exists at most once
 Valid integer inputs
Limitations:
 Array length up to 100,000
 Array values from 109 to 109
 Must run in O(log n) time
So in summary, the scope is finding the range boundaries of a target integer in a sorted array, given assumptions like no duplicates, while meeting time and space complexity constraints.
Here are some ways we can establish boundaries for this binary search problem:
Input Space Boundary:
 Array of integers
 Array is sorted in nondecreasing order
 Array length from 0 to 100,000 elements
 Array values from 109 to 109
 Target value is integer from 109 to 109
Output Space Boundary:
 Start index and end index as integers
 Indices range from 0 to array length  1
 [1, 1] indicates target not found
Algorithm Boundary:
 Must have O(log n) time complexity
 Can leverage binary search algorithm
 Can eliminate half search space each iteration
Objective Boundary:
 Minimize iterations and comparisons
 Do not return actual target elements
 Only need to find range boundaries
By defining boundaries for inputs, outputs, objectives and computational constraints, we can scope the solution space to the most efficient algorithms. This helps eliminate unsuitable approaches.
Distilling the Problem to Its Core Elements
The fundamental concept this problem is based on is efficiently searching for a value in an ordered sequence using binary search. At its core, it is about leveraging the structure of ordered data to optimize lookups.
The simplest way I would describe this problem is:
“Given a sorted list of numbers, quickly find the range of indices that contain a specific target number we’re looking for.”
The core problem is locating a target value range within sorted data. We can simplify the problem statement to:
“Find start and end index of target in a sorted array.”
The key components of the problem are:
 Sorted input array
 Target value to search for
 Start and end indices for target range
The minimal set of operations is:
 Perform binary search on array
 Compare target to middle element
 Eliminate half of search space
 Repeat until range boundaries found
So in summary, this is an optimized search problem that exploits the ordering of data to efficiently hone in on a target value range using binary search. The core idea is rapid lookup in sorted sequences.
Visual Model of the Problem
Here are some ways we could visualize the binary search for range problem statement:
Show the sorted array visually as bars of increasing height. Highlight the target value range.
Animate the binary search process by splitting the array in half each iteration, zooming in on the target range.
Illustrate splitting the array range into lower, middle, and upper segments based on comparisons.
Use a number line marked with array indices and highlight the start and end index range.
Visualize the index range narrowing down over each binary search iteration.
Contrast examples where the target is found vs not found.
Diagram the binary tree implicit in the divide and conquer search process.
Annotate array divisions that are eliminated due to value comparisons.
Show worst case (element not present) and best case (element is middle) pivots.
Display recursion call tree and trace of parameter values in each stack frame.
Leveraging visuals like highlighting, animation, diagrams, and annotations can provide an intuitive sense of how binary search hones in on a target range in a sorted array.
Problem Restatement
Here’s how I would paraphrase the binary search for range problem statement in my own words:
We’re given an array of integers sorted in nondecreasing order. This means each number in the array is less than or equal to the number that comes after it.
We also have a specific target number that we need to search for in the array. Our goal is to find the start and end index in the array that brackets the target number. This will tell us the range of indices that contain the target.
If the target number is not present anywhere in the array, we should return [1, 1] to indicate the target range doesn’t exist.
We need to write an efficient algorithm that can find the target range in O(log n) time, where n is the length of the input array. This suggests using a binary search approach.
The array size can be up to 100,000 elements, and each integer value ranges from 109 to 109, so our algorithm needs to work at scale.
Does this summarize the essence of the problem? Please let me know if I’m misinterpreting any part of the problem statement. Getting the requirements clear upfront will help guide developing the solution.
Abstract Representation of the Problem
Here is one way we could formulate an abstract representation of this binary search for range problem:
Given:
 S, a sorted sequence of n abstract elements
 v, a target value
Objective:
Find the subsequence range of S that encapsulates all occurrences of v based on the following rules:
 Each element e in S has an associated value val(e) that can be compared to v
 If val(e) == v for some element e, then e contains an occurrence of v
 The subsequence range is bounded by the first and last elements e where val(e) == v
Output:
 [i, j] where i and j are the indices of the first and last v elements
 [1, 1] if v does not occur in S
Constraints:
 The range must be found in O(log n) time complexity
By representing the input array abstractly as a sorted sequence, we can focus on the key relationships, objective and constraints without programming details:
 Sorted ordering enables binary search
 Objective is finding bounding indices of value occurrences
 Log n time requirement drives algorithm choices
Please let me know if I’m missing anything important in this abstract representation!
Terminology
Here are some key technical terms relevant to this binary search for range problem:
Binary search  An algorithm for efficiently locating an element in a sorted array by repeatedly dividing the search space in half. Key to the O(log n) solution.
Sorted array  Array where elements are arranged in increasing order. Allows binary search and range identification.
Logarithmic time  An algorithm with O(log n) time divides the problem in half each iteration. Binary search achieves this.
Start/end indices  The first and last positions bounding target value occurrences in the array. Define the range.
Divide and conquer  Breaking down a problem into smaller recursive subproblems. How binary search splits the array.
Ordered data  Data with a meaningful sequential ordering relationship. Enables techniques like binary search.
Range query  Finding the span or boundaries of a target value rather than just existence. Alters binary search.
The key terms like binary search, logarithmic time, divide and conquer, ordered data, and range query suggest applying binary search on the sorted input to efficiently find start and end indices. The terms provide clues to mapping the problem to a solution.
Problem Simplification and Explanation
Here’s one way we can break down this binary search for range problem into simpler concepts and provide an analogy:
At its core, this problem involves:
 Searching  finding a value within a set of data
 Sorted data  data with a meaningful sequential order
 Divide and conquer  breaking down a problem into smaller pieces
We can think of it like searching for a word in a glossary or index. The glossary words are alphabetically sorted.
To look up a word efficiently, we open the book near the middle, check if our target word comes before or after the midpoint, then repeat focusing just on the half where our word could be.
With each comparison, we divide the remaining search space in half  just like binary search divides a sorted array in half each iteration.
Very quickly we hone in on the target word’s position, or determine it’s not present. This logarithmic search process mirrors the problem constraints.
The key ideas are rapidly narrowing down options based on ordering relationships and incremental divide and conquer. Let me know if this highlevel analogy helps explain the core concepts!
Constraints
Here are some specific characteristics of the binary search for range problem that we can leverage:
Array sorted in nondecreasing order  This ordered structure enables binary search to quickly hone in on target location.
Integer value range from 109 to 109  Allows pruning checks when comparing indices in the middle.
Output is index range, not actual elements  Simplifies implementation to just keeping track of indices.
Target may not exist  Can terminate early if search range is exhausted.
Elements unique  No repeated values, so we can stop search when target found.
Array length up to 105  Implies need for O(log n) complexity.
Range query, not just existence  Must adapt binary search to find first and last indices.
Logarithmic time requirement  Points to binary or exponential search specifically.
The ordered integers, range based output, uniqueness, array size, and time requirement all suggest optimizing with an adapted binary search approach. The constraints help narrow down the solution search space.
Here are some key insights gained by analyzing the constraints for this binary search for range problem:
Ordered array enables binary search for efficient lookup.
Large input size implies need for logarithmic complexity.
Integer constraints allow pruning based on value comparisons.
Not returning elements simplifies to just tracking indices.
Bounded range allows defining termination conditions.
No duplicates means we can terminate search early.
Range query requires adapting standard binary search.
Logarithmic time requirement points to binary/exponential search.
Input size up to 105 elements rules out linear search.
Sorted order allows skipping chunks when not found.
The ordered structure, time requirement, lack of duplicates, integer ranges, and rangebased output all indicate binary search is the intended solution. The constraints help derive optimizations and narrow the solution options.
Case Analysis
Here are some additional test cases covering a range of scenarios:
 Target at center of small array:
Input: nums = [2,5,7,9,10], target = 7 Output: [2,2]
Analysis: Basic case, target found at direct center.
 Target at end of large array:
Input: nums = [1,…,1000000], target = 1000000 Output: [99999, 99999]
Analysis: Target at extreme end of wide range, checks edge case.
 Target outside bounds of array:
Input: nums = [1,3,5], target = 0 Output: [1,1]
Analysis: Invalid target, checks not found case.
 Repeated target occurrences:
Input: nums = [2,3,3,3,5], target = 3 Output: [1,3]
Analysis: Multiple instances of target, validates range.
 Empty input array:
Input: nums = [], target = 5
Output: [1,1]
Analysis: Empty array edge case.
 Single element array:
Input: nums = [5], target = 5 Output: [0,0]
Analysis: Singleton array boundary case.
These test a wide array of parameters, bounds, not found cases, repeats, and edge cases to validate the robustness and correctness of the solution.
Here’s one way to categorize test cases for the binary search for range problem:
Basic Cases
 Target at center
 Target at end
Edge Cases
 Empty array
 Singleton array
 Out of bounds target
Target Position
 Center
 End
 Out of bounds
Array Size
 Small
 Large
Target Occurrences
 Unique
 Repeated
The key edge cases are:
 Empty array
 Singleton array
 Target outside array bounds
These test boundary behaviors of the solution by violating normal assumptions.
The other categories like array size and target position cover a range of mainstream and corner case scenarios to thoroughly exercise the implementation.
Good test cases explore both typical and edge cases across multiple dimensions like input parameters, data ranges, and output formats.
Here are some key insights gained by analyzing the different test cases:
Basic cases validate correct functionality on wellbehaved inputs.
Edge cases like empty/singleton arrays reveal boundary behaviors.
Out of bounds targets check not found logic.
Repeated targets highlight properly identifying range.
Varying target positions test search over the full array.
Large inputs evaluate algorithm scalability.
Different output formats (indices vs 1) exercise code branches.
Need to verify closed interval bounds [i,j] vs half open [i, j).
Repeats check continuing search until range fully bounded.
Extremes like center and end targets verify pivoting logic.
Negative numbers test signed integer comparison logic.
Together the cases help ensure the solution works correctly across the full input and output spaces, while revealing potential edge case bugs. They provide assurance of robustness.
Identification of Applicable Theoretical Concepts
Here are some mathematical and algorithmic concepts that could help optimize solving this binary search for range problem:
Binary search  This classic algorithm allows efficiently locating a target value in O(log n) time given a sorted array. Directly applicable.
Divide and conquer  Breaking down the problem into smaller subproblems is embodied in the binary search partitioning of the array.
Logarithmic complexity  Binary search achieves O(log n) search times making it suitable for large n.
Ordered/sorted data structures  The sorted input allows techniques like binary search, skipping, elimination.
Pruning  Range comparisons during search can eliminate portions of indices to reduce search space.
Recursion  Binary search can be implemented recursively, dividing on each function call.
Branch and bound  Establishing bounds on the range output can help prune.
Cache/Memoization  Caching redundant subproblem results could optimize repeated lookups.
Applying concepts like binary search, divide and conquer, recursion, pruning, and exploiting order and structure can help make this problem much more tractable. The constraints align well with characteristics enabling these techniques.
Simple Explanation
Here’s how I would explain this binary search for range problem in simple, nontechnical terms:
Imagine you have a giant phone book with thousands of pages of alphabetically sorted names. I want to look up my friend Alice Abernathy and see what page range her name appears on.
The obvious way would be to start at the first page and flip through one by one until you find the start of the “A” names section, then keep going until you reach the end of the A’s. But that could take forever in a huge phone book!
Instead, you could open up right to the middle page, see if “A” comes before or after that midpoint, then ignore the half of the book that doesn’t apply. Repeat this again and again, zeroing in closer each time. Very quickly you can find the target pages.
This faster search process works because the data is sorted alphabetically. It lets you rule out large portions on every comparison rather than checking each page one at a time. This is the basic idea behind binary search on sorted data!
For a child, I could demonstrate with a physical dictionary or phone book to search for a word in order to explain the concept intuitively. The key idea is honing in on a target efficiently by leveraging ordered structure.
Problem Breakdown and Solution Methodology
Here is how I would approach solving this binary search for range problem:
Overview: The overall process is to leverage a modified binary search to efficiently find the start and end indices of the target value in the sorted array.
Steps:
Initialize left and right boundaries for the search range.
While the left pointer is less than the right:
Calculate the middle index between left and right pointers.
Compare middle element to target.
If equal, expand outwards from middle to set new boundaries.
Else if less, set new left to middle+1 (ignore left half).
Else if greater, set new right to middle1 (ignore right half).
Return final left and right pointers which will mark target range.
If range invalid (left > right), return [1, 1]
This divides the sorted array in half each iteration, eliminating portions that cannot contain the target until the boundaries are found.
Example:
Input: [1,2,5,5,8,9], target = 5
left = 0, right = 5
mid = 2, nums[2] = 5 > left = 2, right = 2
mid = 2, nums[3] = 5 > left = 2, right = 3
Return [2, 3]
Changes like a different search algorithm or added constraints may alter performance but not change the highlevel approach.
Inference of ProblemSolving Approach from the Problem Statement
Here are some key terms and how they guide my solution approach:
Sorted array  The sorted order allows using binary search to efficiently narrow target location.
Logarithmic time  This requirement implies using a logarithmic algorithm like binary search.
Start/end indices  Finding a range instead of just existence means adapting binary search.
Integer constraints  The integer nature allows pruning based on value comparisons.
Divide and conquer  Binary search exemplifies this, breaking the problem into smaller pieces.
Ordered data  Sorted structure enables techniques like binary search, skipping, eliminating ranges.
Range query  Finding the span of the target rather than just existence guides the algorithm.
These terms like sorted data, logarithmic time, range query, divide and conquer, and integer constraints strongly suggest applying a binary search approach. The keywords help map the problem statement onto an optimal algorithm category and adaptation to handle ranges.
How did you infer from the problem statement that this problem can be solved using ?
Simple Explanation of the Proof
I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?
Stepwise Refinement
Here is one way we could break down this binary search for range problem into more refined steps:
 Highlevel approach:
 Use binary search to find target range
 More granular steps:
 Initialize left and right pointers to bounds of array
 While left < right:
 Calculate mid index between left and right
 Compare nums[mid] to target
 If equal, expand outwards from mid
 Else if less, set left = mid + 1
 Else set right = mid  1
 Return final left and right pointers
 Independent parts:
 Calculating mid index
 Comparing element at mid vs target
 Determining new search range based on comparison
 Repeatable patterns:
 Narrowing search range based on mid element comparison
 Halving the search space each iteration
 Moving left/right pointers based on value relative to target
The key is to divide the approach into smaller specific steps, recognize reusable subtasks, and identify the recurring patterns of bisecting the range and narrowing in based on comparisons.
Solution Approach and Analysis
Here is how I would approach solving this binary search for range problem:
Overview: The overall process is to use a modified binary search to efficiently find the start and end index of the target value in the sorted array.
Steps:
Initialize left and right pointers to the bounds of the array.
While left pointer is less than right:
Calculate the mid index between left and right.
Compare value at mid index to target.
If equal, expand outwards from mid to set new range boundaries.
Else if less, set left pointer to mid + 1 to discard left half.
Else set right pointer to mid  1 to discard right half.
Return final left and right pointers which mark target range.
If range invalid (left > right), return [1,1].
This iteratively halves the search space based on comparing array midpoint to target, until range is found.
Example:
Input: [1,3,5,7,8,9], target = 7
left = 0, right = 5
mid = 2, nums[2] = 5 < 7, left = 3
mid = 4, nums[4] = 7 == 7, expand range
Return [3, 4]
Modifications like different data structures or added constraints would change optimal algorithms but not the highlevel approach.
Identify Invariant
One invariant in this binary search for range problem is:
 The target, if present, must be located between the left and right pointers at each step.
This holds true because:
The array is sorted, so elements <= target are to the left, and >= are to the right.
The left and right pointers start at array bounds and converge.
Each comparison discards half the elements, keeping the target range.
We expand the pointers when mid equals target.
This invariant allows us to:
Efficiently halve the search space each iteration.
Know the target can’t be outside current range.
Terminate when range is empty as target not present.
Maintain loop correctness.
By leveraging this loop invariant about the target location, we optimize the binary search process to eliminate portions of the array guaranteed not to contain the target until the range is found.
Identify Loop Invariant
What is the loop invariant in this problem?
Thought Process
Here is how I would think through solving this binary search for range problem:
The problem statement cues:
 Sorted array input
 Target value to search for
 Want start and end indices
 Must be O(log n) time
This suggests:
 Leverage binary search for efficient lookup in ordered data
 Adapt to find range instead of just existence check
 Use pointers to track indices instead of elements
My thinking process:
 Recognize sorted input enables binary search
 Logarithmic time requirement implies binary search
 Range query means adapting standard binary search
 Use left and right pointers to track range
 Iterate halving search space based on comparisons
 Expand pointers when midpoint equals target
 Return final pointer values as range
Python code:


The key is recognizing the constraints imply binary search, and adapting it to track range.
Here is one way to complete the binary search for range Python code:


The key steps are:
 Initialize left/right pointers
 Binary search while loop
 Expand pointers when target found
 Return final range
 Return [1,1] if not found
Establishing Preconditions and Postconditions
Here are the details for the searchRange binary search method:
 Parameters:
 nums: List[int]  Sorted input array of integers
 target: int  Target value to search for
 Preconditions:
 nums is sorted in ascending order
 target is an integer that may or may not be in nums
 Method Functionality:
 Performs binary search on nums looking for start/end indices of target
 Returns a list [left, right] with target range
 Postconditions:
 Returns [left, right] target range indices
 Returns [1, 1] if target not found
 Does not modify original nums array
 Error Handling:
 If nums is not sorted, behavior is undefined
 Could throw exception or fail silently with incorrect result
 If target is not numeric, could cast or throw exception
Let me know if you need any clarification or have additional questions!
Problem Decomposition
You’ve outlined a very structured approach to decomposing a problem into modular, reusable components. Breaking down complex problems into coherent subtasks and identifying opportunities for abstraction are essential skills in programming. However, without a specific problem statement to analyze, it’s difficult for me to provide concrete details for each of those steps.
To demonstrate meaningful problem decomposition, I would need some initial context, such as:
 A highlevel description of the overall problem
 Defined inputs, outputs, goals
 Key constraints or requirements
 Expected size or complexity
With those problem specifics, I could apply this structured analysis to:
 Explain the problem in my own words
 Identify major subproblems
 Break those down further into refined tasks
 Recognize reusable patterns and extract into functions
 Name the functions appropriately
 Illustrate dependencies and interactions
This would lead to a modularized, abstracted breakdown of the solution space into coherent pieces. Please feel free to provide some initial problem context and I can try walking through this decomposition process in a concrete way. Having that specific starting point will help demonstrate each of the steps you have outlined.
From Brute Force to Optimal Solution
Here is one approach to illustrating a brute force solution for the binary search for range problem and then optimizing it:
Brute force solution:


This scans linearly through the array comparing each element to the target. If found, it expands outwards to find the full range.
 Time complexity: O(n)
 Space complexity: O(1)
Optimized solution:


This uses binary search to efficiently hone in on the target location rather than linear scan.
 Time complexity: O(log n)
 Space complexity: O(1)
So converting from brute force linear scan to binary search provides a major optimization, improving time complexity from O(n) to O(log n). The iterative implementation also maintains constant space usage.
Code Explanation and Design Decisions
To provide a meaningful explanation covering those key elements, I would need some initial context about the specific problem and associated code. Without details on the problem statement, expected solution, and example code, it’s difficult to analyze the intent behind parameters, loop logic, conditions, updates, invariants, and outputs.
However, I’m happy to go through that process for a specific piece of code! Some examples of helpful context would include:
The overall problem being solved
What class or component this method is part of
The method name and signature
Expected parameters and return types
Assumptions about program state
With those specifics I could better characterize the role and behavior of the method in the broader context. Please feel free to provide any additional background details that may help drive a more detailed and structural analysis.
Coding Constructs
Based on the binary search code provided:
The highlevel problem solving strategies appear to be binary search and divideandconquer. The code leverages binary search to efficiently lookup a target value in a sorted array.
I would explain this code as implementing a fast search algorithm that quickly finds the location of a value in an ordered list by repeatedly dividing the search space in half.
The logical constructs used are a while loop, ifelif conditional branching, variable assignment, and boolean comparisons. These build up the binary search algorithm independent of language syntax.
In plain English, the algorithm approaches the problem by first initializing bounds for the search space. It then loops, calculating a mid point, comparing the mid value to the target, and adjusting the bounds based on that comparison to narrow in on the target.
The key steps are: initializing left/right bounds, calculating mid index, comparing mid element, updating left/right bounds based on comparison, and returning final position. This quickly hones in on the target location by repeatedly halving the search space.
The general patterns are binary search and divideandconquer. It exemplifies bisecting an ordered data set to optimally find an element, irrespective of programming language used.
Language Agnostic Coding Drills
Key Coding Concepts:
Boundary initialization  Setting initial values for variables like search bounds. (Easy)
Looping  Using a loop structure to repeat a process. (Easy)
Conditional logic  Branching code execution based on comparing values. (Intermediate)
Calculating midpoint  Arithmetic algorithms like finding middle index. (Easy)
Search space pruning  Eliminating sets of values from consideration. (Advanced)
Range expansion  Extending boundaries once target found. (Intermediate)
ProblemSolving Approach:
Initialize left/right bounds to define search space
Use loop to repeatedly narrow down search range
Calculate mid index and fetch element
Apply conditional logic to compare element to target
Prune away half of search space based on comparison
When target found, expand range outwards
Return final indices once range fully established
The key is starting with simple concepts like variables and looping, then progressively building up to advanced techniques like pruning. The stepbystep approach mirrors this increasing complexity to arrive at the final optimized binary search solution. Please let me know if this helps explain the concept breakdown and strategy in a clearer manner!
Targeted Drills in Python
 Boundary initialization:


 Looping:


 Midpoint calculation:


 Conditionals:


 Search space pruning:


These would be assembled in order  initialize bounds, loop, calculate mid, use conditionals to compare and prune search space, and expand range once target found.
The key problemspecific concept is expanding the left/right bounds when target is found until boundaries are reached. This expanding range step builds on the core concepts like looping and conditionals.
Q&A
Similar Problems
Can you suggest 10 problems from LeetCode that require similar problemsolving strategies or use similar underlying concepts as the problem we’ve just solved? These problems can be from any domain or topic, but they should involve similar steps or techniques in the solution process. Also, please briefly explain why you consider each of these problems to be related to our original problem.