public class Solution extends Object
2161 - Partition Array According to Given Pivot.
Medium
You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that the following conditions are satisfied:
pivot appears before every element greater than pivot.pivot appears in between the elements less than and greater than pivot.pivot and the elements greater than pivot is maintained.
pi, pj where pi is the new position of the ith element and pj is the new position of the jth element. For elements less than pivot, if i < j and nums[i] < pivot and nums[j] < pivot, then pi < pj. Similarly for elements greater than pivot, if i < j and nums[i] > pivot and nums[j] > pivot, then pi < pj.Return nums after the rearrangement.
Example 1:
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.
Example 2:
Input: nums = [-3,4,3,2], pivot = 2
Output: [-3,2,4,3]
Explanation: The element -3 is less than the pivot so it is on the left side of the array.
elements 4 and 3 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.
[-3] and [4, 3] are the respective orderings.
Constraints:
1 <= nums.length <= 105-106 <= nums[i] <= 106pivot equals to an element of nums.| Constructor and Description |
|---|
Solution() |
| Modifier and Type | Method and Description |
|---|---|
int[] |
pivotArray(int[] nums,
int pivot) |
Copyright © 2022. All rights reserved.