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

Difficulty Level : Medium
Last Updated : 20 Oct, 2020

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-a) For each element x in arr[k] to arr[n-1]. O(n-k)
If x is greater than the min then remove min from temp[] and insert x.
3-b)Then, determine the new min from temp[]. O(k)
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 the above.

C++

// 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

// 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

Python

''' 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.

C#

// C# code for k largest elements in an array
using System;
 
class GFG {
    public static void kLargest(int[] 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
        Array.Sort(arr);
        Array.Reverse(arr);
 
        // Print the first kth largest elements
        for (int i = 0; i < k; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = new int[] { 1, 23, 12, 9,
                                30, 2, 50 };
        int k = 3;
        kLargest(arr, k);
    }
}
 
// This code contributed by Rajput-Ji

PHP

<?php 
// PHP code for k largest 
// elements in an array
 
function kLargest(&$arr, $n, $k)
{
    // Sort the given array arr 
    // in reverse order.
    rsort($arr);
 
    // Print the first kth 
    // largest elements
    for ($i = 0; $i < $k; $i++)
        echo $arr[$i] . " ";
}
 
// Driver Code
$arr = array(1, 23, 12, 9, 
                30, 2, 50);
$n = sizeof($arr);
$k = 3;
kLargest($arr, $n, $k);
 
// This code is contributed 
// by ChitraNayal
?>

Output:

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.

C++

#include <iostream>
using namespace std;
 
// Swap function to interchange
// the value of variables x and y
int swap(int& x, int& y)
{
    int temp = x;
    x = y;
    y = temp;
}
 
// Min Heap Class
// arr holds reference to an integer 
// array size indicate the number of
// elements in Min Heap
class MinHeap {
 
    int size;
    int* arr;
 
public:
    // Constructor to initialize the size and arr
    MinHeap(int size, int input[]);
 
    // Min Heapify function, that assumes that
    // 2*i+1 and 2*i+2 are min heap and fix the
    // heap property for i.
    void heapify(int i);
 
    // Build the min heap, by calling heapify
    // for all non-leaf nodes.
    void buildHeap();
};
 
// Constructor to initialize data
// members and creating mean heap
MinHeap::MinHeap(int size, int input[])
{
    // Initializing arr and size
 
    this->size = size;
    this->arr = input;
 
    // Building the Min Heap
    buildHeap();
}
 
// Min Heapify function, that assumes
// 2*i+1 and 2*i+2 are min heap and 
// fix min heap property for i
 
void MinHeap::heapify(int i)
{
    // If Leaf Node, Simply return
    if (i >= size / 2)
        return;
 
    // variable to store the smallest element
    // index out of i, 2*i+1 and 2*i+2
    int smallest;
 
    // Index of left node
    int left = 2 * i + 1;
 
    // Index of right node
    int right = 2 * i + 2;
 
    // Select minimum from left node and 
    // current node i, and store the minimum
    // index in smallest variable
    smallest = arr[left] < arr[i] ? left : i;
 
    // If right child exist, compare and
    // update the smallest variable
    if (right < size)
        smallest = arr[right] < arr[smallest]
                             ? right : smallest;
 
    // If Node i violates the min heap 
    // property, swap  current node i with
    // smallest to fix the min-heap property 
    // and recursively call heapify for node smallest.
    if (smallest != i) {
        swap(arr[i], arr[smallest]);
        heapify(smallest);
    }
}
 
// Build Min Heap
void MinHeap::buildHeap()
{
    // Calling Heapify for all non leaf nodes
    for (int i = size / 2 - 1; i >= 0; i--) {
        heapify(i);
    }
}
 
void FirstKelements(int arr[],int size,int k){
    // Creating Min Heap for given
    // array with only k elements
    MinHeap* m = new MinHeap(k, arr);
 
    // Loop For each element in array
    // after the kth element
    for (int i = k; i < size; i++) {
 
        // if current element is smaller 
        // than minimum element, do nothing 
        // and continue to next element
        if (arr[0] > arr[i])
            continue;
 
        // Otherwise Change minimum element to 
        // current element, and call heapify to
        // restore the heap property
        else {
            arr[0] = arr[i];
            m->heapify(0);
        }
    }
    // Now min heap contains k maximum
    // elements, Iterate and print
    for (int i = 0; i < k; i++) {
        cout << arr[i] << " ";
    }
}
// Driver Program
int main()
{
 
    int arr[] = { 11, 3, 2, 1, 15, 5, 4,
                           45, 88, 96, 50, 45 };
 
    int size = sizeof(arr) / sizeof(arr[0]);
 
    // Size of Min Heap
    int k = 3;
 
    FirstKelements(arr,size,k);
 
    return 0;
}
// This code is contributed by Ankur Goel

Java

import java.io.*;
import java.util.*;
 
class GFG{
   
public static void FirstKelements(int arr[], 
                                  int size, 
                                  int k)
{
     
    // Creating Min Heap for given
    // array with only k elements
    // Create min heap with priority queue
    PriorityQueue<Integer> minHeap = new PriorityQueue<>();
    for(int i = 0; i < k; i++)
    {
        minHeap.add(arr[i]);
    }
     
    // Loop For each element in array
    // after the kth element
    for(int i = k; i < size; i++)
    {
         
        // If current element is smaller 
        // than minimum ((top element of
        // the minHeap) element, do nothing 
        // and continue to next element
        if (minHeap.peek() > arr[i])
            continue;
         
        // Otherwise Change minimum element
        // (top element of the minHeap) to 
        // current element by polling out 
        // the top element of the minHeap
        else
        {
            minHeap.poll();
            minHeap.add(arr[i]);
        }
    }
     
    // Now min heap contains k maximum
    // elements, Iterate and print
    Iterator iterator = minHeap.iterator(); 
     
    while (iterator.hasNext()) 
    { 
        System.out.print(iterator.next() + " "); 
    } 
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 11, 3, 2, 1, 15, 5, 4,
                  45, 88, 96, 50, 45 };
     
    int size = arr.length;
     
    // Size of Min Heap
    int k = 3;
     
    FirstKelements(arr, size, k);
}
}
 
// This code is contributed by Vansh Sethi

Output

50 88 96

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

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