## DEV Community

ashutosh049

Posted on • Updated on

# Kth Largest Element in an Array

You can refer to the Leetcode problem 215. Kth Largest Element in an Array

#### Problem Statement

Given an integer array `nums` and an integer `k`, return the `kth` largest element in the array.

Note that it is the `kth` largest element in the sorted order, not the kth distinct element.

### Example

``````Input: nums = [3,2,1,5,6,4], k = 2
Output: 5
``````

### Approach 1: Using Sorting

We can use sorting to first sort the nums array in natural order and then return kth element from end i.e n-k th element from start.

K index from end is equal to length-k index from start

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

#### Complexity Analysis

TC: `O(N log N)`, where `N` is the size of the input array
SC: `O(1)`

### Approach 2: Using Heap

Actually there are multiple sub-approaches if we choose to use `Heap`.

Approach Number of elements number of poll
Min Heap of size N Min heap to store all N elements(at most `N-K` minimum elements at any given time) poll for `N-K` times to get `kth`largest
Min Heap of size K Min heap to store at most K elements. Adding new elements after first K elements are added, we check if new element is greater than heap root(peek) and if so we delete it first and then add the new greater element, otherwise not poll for 1 time and return the polled element
Max Heap of size N Max heap to store all N elements(at most `K` maximum elements at any given time) poll for `K` times to get `kth`largest
##### Min Heap of size N
``````class Solution {

public int findKthLargest(int[] nums, int k) {
int n = nums.length;

if (n == 1) {
return nums[0];
}

PriorityQueue<Integer> minHeap = new PriorityQueue();

for(int num: nums){
minHeap.offer(num);
}

int i = 0;
int kThLargest = minHeap.poll();

while (i++ < (n - k)) {
kThLargest = minHeap.poll();
}

return kThLargest;
}
}
``````
##### Min Heap of size K
``````import java.util.Arrays;
import java.util.Collections;

//Min Heap of size K
class Solution {

public int findKthLargest(int[] nums, int k) {
int n = nums.length;

if (n == 1) {
return nums[0];
}

PriorityQueue<Integer> minHeap = new PriorityQueue(k);

for(int i=0; i<k; i++){
minHeap.offer(nums[i]);
}

for(int i=k; i<n; i++){
if(minHeap.peek() < nums[i]){
minHeap.poll();
minHeap.offer(nums[i]);
}
}

return minHeap.poll();
}
}
``````
##### Max Heap of size N
``````class Solution {
public int findKthLargest(int[] nums, int k) {

int len = nums.length;

if(len == 1){
return nums[0];
}

// Since we are using Max-Heap, we need to sort accordingly
Comparator<Integer> comp = (a,b) -> b.compareTo(a);
PriorityQueue<Integer> maxHeap = new PriorityQueue<>(comp);

for(int num: nums){
maxHeap.offer(num);
}

// we need to poll for k-1 times and
// return the next polled element
int i = 1;

while(i++ < k){
maxHeap.poll();
}

return  maxHeap.poll();

}
}
``````

Problem Credit : leetcode.com