# 215 Kth Largest Element in an Array – Medium

### Problem:

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

For example, Given [3,2,1,5,6,4] and k = 2, return 5.

Note: You may assume k is always valid, 1 ≤ k ≤ array’s length.

### Thoughts:

The trivial solution would be using a sort, either quicksort or mergesort.

Since sorting is O(nlogn), and additional step is to turn the nums.length – k element, so this trivial approach is O(nlogn).

Better approach is to use a quickSelect way. Average running time for quickSelect is O(n), so this is a faster approach comparing the sorting approach.

In the solution below, quickSelect it to select the (k+1)th smallest element instead.

### Solutions:

``````public class Solution {
public int findKthLargest(int[] nums, int k) {
return quickSelect(nums, 0, nums.length - 1, k);
}
private int quickSelect(int[] nums, int start, int end, int k) {
int index = partition(nums, start, end, k);
int right = end - index + 1;
if (right == k) {
return nums[index];
}
if (right > k) {
return quickSelect(nums, index + 1, end, k);
}
else {
return quickSelect(nums, start, index - 1, k - right);
}
}
private int partition(int[] nums, int start, int end, int k) {
int pivot = nums[end];
int index = end - 1;
while (start <= index) {
if (nums[start] > pivot) {
while (index > start && nums[index] > pivot) {
index --;
}
int tmp = nums[start];
nums[start] = nums[index];
nums[index] = tmp;
index --;
}
start ++;
}
nums[end] = nums[index + 1];
nums[index + 1] = pivot;
return index + 1;
}

}
``````