LeetCode in Kotlin

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:

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:

Solution

class Solution {
    fun pivotArray(nums: IntArray, pivot: Int): IntArray {
        val ans = IntArray(nums.size)
        var point = 0
        var equal = 0
        for (i in nums) {
            if (i < pivot) {
                ans[point] = i
                ++point
            } else if (i == pivot) {
                ++equal
            }
        }
        while (equal > 0) {
            ans[point] = pivot
            ++point
            --equal
        }
        for (i in nums) {
            if (i > pivot) {
                ans[point] = i
                ++point
            }
        }
        return ans
    }
}