Optimize partitioning for O(n) time
Last updated: September 21, 2025
Quick Overview
Given an array of integers, partition the array into two subsets such that all elements in one subset are less than or equal to a specified pivot value, and all elements in the other subset are greater than the pivot. The solution should achieve this partitioning in O(n) time complexity, returning the rearranged array as the output.
Walmart
September 21, 2025248
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Given an array of integers, partition the array into two subsets such that all elements in one subset are less than or equal to a specified pivot value, and all elements in the other subset are greater than the pivot. The solution should achieve this partitioning in O(n) time complexity, returning the rearranged array as the output.
How to Approach This
- Clarify input constraints and edge cases before writing code.
- Walk through your approach verbally and confirm with the interviewer before coding.
- Start with a brute force solution, then optimize. Mention time and space complexity.
- Test your solution with examples, including edge cases like empty input or duplicates.
- Consider common patterns: sliding window, two pointers, hash map, BFS/DFS, dynamic programming.
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Problem Analysis
The problem requires partitioning an array into two subsets based on a pivot value. This suggests the use of the two pointers approach, where one pointer starts from the beginning of the array (le...
Approach
- Initialize two pointers:
leftat the start (index 0) andrightat the end (index len(array) - 1) of the array. - While
leftis less than or equal toright:- If the element at `array[le...
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