Kth Largest Element in an Array

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

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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

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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)

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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 kthlargest
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 kthlargest

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);

        // Add all the elements
        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