# Partition Array

`Sort`, `Array`, `Two Pointers`

Medium

## Problem

Given an array nums of integers and an int k, partition the array (i.e move the elements in "nums") such that:

• All elements < k are moved to the left
• All elements >= k are moved to the right
• Return the partitioning index, i.e the first index i nums[i] >= k.

Notice

You should do really partition in array nums instead of just counting the numbers of integers smaller than k.

If all elements in nums are smaller than k, then return nums.length

## Solution

Quick Select

``````public class Solution {
private void swap(int i, int j, int[] arr) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
/**
*@param nums: The integer array you should partition
*@param k: As description
*return: The index after partition
*/
public int partitionArray(int[] nums, int k) {

int pl = 0;
int pr = nums.length - 1;
while (pl <= pr) {
while (pl <= pr && nums[pl] < k) {
pl++;
}
while (pl <= pr && nums[pr] >= k) {
pr--;
}
if (pl <= pr) {
swap(pl, pr, nums);
}
}
return pl;
}
}
``````

Another approach: O(n)

``````public class Solution {
/**
* @param nums: The integer array you should partition
* @param k: An integer
* @return: The index after partition
*/
public int partitionArray(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}

int offset = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] < k) {
int temp = nums[i];
nums[i] = nums[offset];
nums[offset] = temp;
offset ++;
}
}

return offset;
}
}
``````