# 4ms java solution | Concise Randomized Quick Select implemention

• ``````public class Solution {

public int findKthLargest(int[] nums, int k) {
if (nums == null || k <= 0 || k > nums.length)
return -1;
int i = 0, j = nums.length - 1;
while (true) {
int pivot = partition(nums, i, j);
if (pivot + 1 == k)
return nums[pivot];
if (pivot + 1 < k)
i = pivot + 1;
else
j = pivot - 1;
}
}

private int partition(int[] a, int start, int end) {
swap(a, (int) (Math.random() * (end - start + 1)) + start, end);
int i = start;
for (int j = start; j < end; j++) {
if (a[j] > a[end])
swap(a, i++, j);
}
swap(a, i, end);
return i;
}

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

}

• 4ms, 2 lines of Java

``````public class Solution {
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);

return nums[nums.length -1 -(k-1)];
}
}``````

• Its correct but time complexity of your solution is O(nlogn) + O(logn) stack space assuming the sort method uses merge sort. The QuickSelect Algorithm allows you to find kth largest number on O(n) amortized time.

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