# Beats 95% java submission 2MS solution with explanation

• The trick is in modified quick sort. Condition is not to sort it further.

``````public class Solution {
private int[] array = null;
private int K = 0;
public int findKthLargest(int[] nums, int k) {
k = nums.length - k;
int lo = 0;
int hi = nums.length - 1;
array = nums;
K = k;
quickSort(lo, hi);
return nums[k];
}

private void quickSort(int lowerIndex, int higherIndex) {

int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
// Divide into two arrays
while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j) {
//if upper bound is in KPosition sort else no sorting needed
if(K <= j) {
quickSort(lowerIndex, j);
}
}
if (i < higherIndex) {
//if lower bound is in KPosition sort else no sorting needed
if(K >= i) {
quickSort(i, higherIndex);
}
}
}

private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
``````

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