Leetcode #2161: Partition Array According to Given Pivot
In this guide, we solve Leetcode #2161 Partition Array According to Given Pivot in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that the following conditions are satisfied: Every element less than pivot appears before every element greater than pivot.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Two Pointers, Simulation
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: nums = [9,12,5,10,14,3,10], pivot = 10
Output: [9,5,3,10,10,12,14]
Explanation:
The elements 9, 5, and 3 are less than the pivot so they are on the left side of the array.
The elements 12 and 14 are greater than the pivot so they are on the right side of the array.
The relative ordering of the elements less than and greater than pivot is also maintained. [9, 5, 3] and [12, 14] are the respective orderings.
Python Solution
class Solution:
def pivotArray(self, nums: List[int], pivot: int) -> List[int]:
a, b, c = [], [], []
for x in nums:
if x < pivot:
a.append(x)
elif x == pivot:
b.append(x)
else:
c.append(x)
return a + b + c
Complexity
The time complexity is O(n) (after optional sort O(n log n)). The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.