k largest(or smallest) elements in an array | added Min Heap method

Question: Write an efficient program for printing k largest elements in an array. Elements in array can be in any order.

For example, if given array is [1, 23, 12, 9, 30, 2, 50] and you are asked for the largest 3 elements i.e., k = 3 then your program should print 50, 30 and 23.

Method 1 (Use Bubble k times)
Thanks to Shailendra for suggesting this approach.
1) Modify Bubble Sort to run the outer loop at most k times.
2) Print the last k elements of the array obtained in step 1.

Time Complexity: O(nk)

Like Bubble sort, other sorting algorithms like Selection Sort can also be modified to get the k largest elements.

Method 2 (Use temporary array)
K largest elements from arr[0..n-1]

1) Store the first k elements in a temporary array temp[0..k-1].
2) Find the smallest element in temp[], let the smallest element be min.
3) For each element x in arr[k] to arr[n-1]
If x is greater than the min then remove min from temp[] and insert x.
4) Print final k elements of temp[]

Time Complexity: O((n-k)*k). If we want the output sorted then O((n-k)*k + klogk)

Thanks to nesamani1822 for suggesting this method.

Method 3(Use Sorting)
1) Sort the elements in descending order in O(nLogn)
2) Print the first k numbers of the sorted array O(k).
Following is the implementation of above.

// C++ code for k largest elements in an array
#include<bits/stdc++.h>
using namespace std;

void kLargest(int arr[], int n, int k)
{
    // Sort the given array arr in reverse 
    // order.
    sort(arr, arr+n, greater<int>());

    // Print the first kth largest elements
    for (int i = 0; i < k; i++)
        cout << arr[i] << " ";
}

// driver program
int main()
{
    int arr[] = {1, 23, 12, 9, 30, 2, 50};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 3;
    kLargest(arr, n, k);
}

// This article is contributed by Chhavi

// Java code for k largest elements in an array
import java.util.Arrays;
import java.util.Collections;

class GFG
{
    public static void kLargest(Integer [] arr, int k) 
    {
        // Sort the given array arr in reverse order
        // This method doesn't work with primitive data
        // types. So, instead of int, Integer type 
        // array will be used 
        Arrays.sort(arr, Collections.reverseOrder()); 
    
    
    // Print the first kth largest elements 
    for (int i = 0; i < k; i++) 
    System.out.print(arr[i] + " ");
    } 
    
    
    public static void main(String[] args) 
    {
        Integer arr[] = new Integer[]{1, 23, 12, 9, 
                                       30, 2, 50};
        int k = 3;
        kLargest(arr,k);     
    }
}
// This code is contributed by Kamal Rawal

''' Python3 code for k largest elements in an array'''

def kLargest(arr, k):
    # Sort the given array arr in reverse 
    # order.
    arr.sort(reverse=True)
    #Print the first kth largest elements
    for i in range(k):
        print (arr[i],end=" ") 

# Driver program
arr = [1, 23, 12, 9, 30, 2, 50]
#n = len(arr)
k = 3
kLargest(arr, k)

# This code is contributed by shreyanshi_arun.

50 30 23 

Time complexity: O(nlogn)

Method 4 (Use Max Heap)
1) Build a Max Heap tree in O(n)
2) Use Extract Max k times to get k maximum elements from the Max Heap O(klogn)

Time complexity: O(n + klogn)

Method 5(Use Oder Statistics)
1) Use order statistic algorithm to find the kth largest element. Please see the topic selection in worst-case linear time O(n)
2) Use QuickSort Partition algorithm to partition around the kth largest number O(n).
3) Sort the k-1 elements (elements greater than the kth largest element) O(kLogk). This step is needed only if sorted output is required.

Time complexity: O(n) if we don’t need the sorted output, otherwise O(n+kLogk)

Thanks to Shilpi for suggesting the first two approaches.

Method 6 (Use Min Heap)
This method is mainly an optimization of method 1. Instead of using temp[] array, use Min Heap.

1) Build a Min Heap MH of the first k elements (arr[0] to arr[k-1]) of the given array. O(k)

2) For each element, after the kth element (arr[k] to arr[n-1]), compare it with root of MH.
……a) If the element is greater than the root then make it root and call heapify for MH
……b) Else ignore it.
// The step 2 is O((n-k)*logk)

3) Finally, MH has k largest elements and root of the MH is the kth largest element.

Time Complexity: O(k + (n-k)Logk) without sorted output. If sorted output is needed then O(k + (n-k)Logk + kLogk)

All of the above methods can also be used to find the kth largest (or smallest) element.

Please write comments if you find any of the above explanations/algorithms incorrect, or find better ways to solve the same problem.

References:
http://en.wikipedia.org/wiki/Selection_algorithm