Find k maximum elements of array in original order

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Given an array arr[] and an integer k, we need to print k maximum elements of given array. The elements should printed in the order of the input.
Note : k is always less than or equal to n.

Examples:

Input : arr[] = {10 50 30 60 15}
        k = 2
Output : 50 60
The top 2 elements are printed
as per their appearance in original
array.

Input : arr[] = {50 8 45 12 25 40 84}
            k = 3
Output : 50 45 84

Method 1: We search for the maximum element k times in the given array. Each time we find one maximum element, we print it and replace it with minus infinite (INT_MIN in C) in the array. Also, the position of all k maximum elements is marked using an array so that with the help of that array we can print the elements in the order given in the original array. The time complexity of this method is O(n*k).

Below is the implementation of the above approach:

C++

// C++ program to find k maximum elements
// of array in original order
#include <bits/stdc++.h>
using namespace std;
 
// Function to print k Maximum elements
void printMax(int arr[], int k, int n)
{
    int brr[n]={0},crr[n];
     
    // Copying the array arr
    // into crr so that it 
    // can be used later
    for(int i=0;i<n;i++)
    {
        crr[i]=arr[i];
    }
    // Iterating for K-times
    for(int i=0;i<k;i++)
    {
        // Finding the maximum element
        // along with its index
        int maxi=INT_MIN;
        int index;
        for(int j=0;j<n;j++)
        {
            if(maxi<arr[j])
            {
                maxi=arr[j];
                index=j;
            }
        }
        // Assigning 1 in order
        // to mark the position 
        // of all k maximum numbers
        brr[index]=1;
        arr[index]=INT_MIN;
    }
     
    for(int i=0;i<n;i++)
    {
        // Printing the k maximum
        // elements array 
        if(brr[i]==1)
        cout<<crr[i]<<" ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 50, 8, 45, 12, 25, 40, 84 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    printMax(arr, k, n);
    return 0;
}
 
// This code is contributed by Pushpesh Raj.

Java

// JAVA program to find k maximum elements
// of array in original order
import java.util.*;
class GFG {
 
  // Function to print k Maximum elements
  public static void printMax(int arr[], int k, int n)
  {
    int brr[] = new int[n];
    int crr[] = new int[n];
    for (int i = 0; i < n; i++)
      brr[i] = 0;
     
    // Copying the array arr
    // into crr so that it
    // can be used later
    for (int i = 0; i < n; i++) {
      crr[i] = arr[i];
    }
     
    // Iterating for K-times
    for (int i = 0; i < k; i++)
    {
       
      // Finding the maximum element
      // along with its index
      int maxi = Integer.MIN_VALUE;
      int index = 0;
      for (int j = 0; j < n; j++) {
        if (maxi < arr[j]) {
          maxi = arr[j];
          index = j;
        }
      }
       
      // Assigning 1 in order
      // to mark the position
      // of all k maximum numbers
      brr[index] = 1;
      arr[index] = Integer.MIN_VALUE;
    }
 
    for (int i = 0; i < n; i++)
    {
       
      // Printing the k maximum
      // elements array
      if (brr[i] == 1)
        System.out.print(crr[i] + " ");
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = new int[] { 50, 8, 45, 12, 25, 40, 84 };
    int n = arr.length;
    int k = 3;
    printMax(arr, k, n);
  }
}
 
// This code is contributed by Taranpreet

Python3

# Function to print k Maximum elements
def printMax(arr, k, n):
    brr = [0 for _ in range(n)]
    crr = [0 for _ in range(n)]
 
    # Copying the array arr
    # into crr so that it
    # can be used later
    for i in range(0, n):
        crr[i] = arr[i]
         
    # Iterating for K-times
    for i in range(0, k):
       
       
        # Finding the maximum element
        # along with its index
        maxi = -99999
        index = 0
        for j in range(0, n):
            if maxi < arr[j]:
                maxi = arr[j]
                index = j
                 
        # Assigning 1 in order
        # to mark the position
        # of all k maximum numbers
        brr[index] = 1
        arr[index] = -99999
 
    for i in range(0, n):
        # Printing the k maximum
        # elements array
        if brr[i] == 1:
            print(crr[i], end='')
            print(" ", end='')
 
if __name__ == "__main__":
    arr = [50, 8, 45, 12, 25, 40, 84]
    n = len(arr)
    k = 3
    printMax(arr, k, n)
 
# This code is contributed by Aarti_Rathi

C#

// C# program to find k maximum
// elements of array in original order
using System;
using System.Linq;
 
class GFG {
 
    // Function to print m Maximum elements
    public static void printMax(int[] arr, int k, int n)
    {
 
        // Copying the array arr
        // into crr so that it
        // can be used later
        int[] brr = new int[n];
        int[] crr = new int[n];
 
        for (int i = 0; i < n; i++) {
            brr[i] = 0;
            crr[i] = arr[i];
        }
 
        // Iterating for K-times
        for (int i = 0; i < k; i++) {
            // Finding the maximum element
            // along with its index
            int maxi = Int32.MinValue;
            int index = 0;
            for (int j = 0; j < n; j++) {
                if (maxi < arr[j]) {
                    maxi = arr[j];
                    index = j;
                }
            }
            // Assigning 1 in order
            // to mark the position
            // of all k maximum numbers
            brr[index] = 1;
            arr[index] = Int32.MinValue;
        }
 
        for (int i = 0; i < n; i++) {
            // Printing the k maximum
            // elements array
            if (brr[i] == 1)
                Console.Write(crr[i] + " ");
        }
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 50, 8, 45, 12, 25, 40, 84 };
        int n = arr.Length;
        int k = 3;
 
        printMax(arr, k, n);
    }
}
// This code is contributed by Aarti_Rathi

Javascript

// Function to print k Maximum elements
 
function printMax(arr, k, n)
{
    var brr = Array(n).fill(0);
    var crr = Array(n).fill(0);
    for (var i =0; i < n; i++)
    {
        brr[i] = 0;
    }
    // Copying the array arr
    // into crr so that it
    // can be used later
    for (var i=0; i < n; i++)
    {
        crr[i] = arr[i];
    }
    // Iterating for K-times
    for (var i=0; i < k; i++)
    {
        // Finding the maximum element
        // along with its index
        var maxi = -Number.MAX_VALUE;
        var index = 0;
        for (var j =0; j < n; j++)
        {
            if (maxi < arr[j])
            {
                maxi = arr[j];
                index = j;
            }
        }
        // Assigning 1 in order
        // to mark the position
        // of all k maximum numbers
        brr[index] = 1;
        arr[index] = -Number.MAX_VALUE;
    }
    for (var i=0; i < n; i++)
    {
        // Printing the k maximum
        // elements array
        if (brr[i] == 1)
        {
            console.log(crr[i] + " ");
        }
    }
}
 
// Driver code
var arr = [50, 8, 45, 12, 25, 40, 84];
var n = arr.length;
var k = 3;
printMax(arr, k, n);
 
// This code is contributed by Aarti_Rathi

Output

50 45 84

Time Complexity: O(n*k)
Auxiliary Space: O(n)

Method 2: In this method, we store the original array in a new array and will sort the new array in descending order. After sorting, we iterate the original array from 0 to n and print all those elements that appear in first k elements of new array. For searching, we can do Binary Search.

C++

// C++ program to find k maximum elements 
// of array in original order
#include <bits/stdc++.h>
using namespace std;
 
// Function to print m Maximum elements
void printMax(int arr[], int k, int n)
{
    // vector to store the copy of the
    // original array
    vector<int> brr(arr, arr + n);
 
    // Sorting the vector in descending
    // order. Please refer below link for
    // details
    // https://www.zambiatek.com/sort-c-stl/
    sort(brr.begin(), brr.end(), greater<int>());
 
    // Traversing through original array and
    // printing all those elements that are
    // in first k of sorted vector.
    // Please refer https://goo.gl/44Rwgt
    // for details of binary_search()
    for (int i = 0; i < n; ++i)
        if (binary_search(brr.begin(),
                 brr.begin() + k, arr[i], 
                        greater<int>()))
            cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 50, 8, 45, 12, 25, 40, 84 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    printMax(arr, k, n);
    return 0;
}

Java

// Java program to find k maximum  
// elements of array in original order
import java.util.Arrays;
import java.util.Collections;
 
public class GfG {
     
    // Function to print m Maximum elements
    public static void printMax(int arr[], int k, int n)
    {
        // Array to store the copy 
        // of the original array
        Integer[] brr = new Integer[n];
         
        for (int i = 0; i < n; i++)
        brr[i] = arr[i];
         
        // Sorting the array in 
        // descending order
        Arrays.sort(brr, Collections.reverseOrder());
     
        // Traversing through original array and
        // printing all those elements that are
        // in first k of sorted array.
        // Please refer https://goo.gl/uj5RCD
        // for details of Arrays.binarySearch()
        for (int i = 0; i < n; ++i)
            if (Arrays.binarySearch(brr, arr[i], 
                    Collections.reverseOrder()) >= 0
                 && Arrays.binarySearch(brr, arr[i], 
                    Collections.reverseOrder()) < k)
                     
                System.out.print(arr[i]+ " ");
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 50, 8, 45, 12, 25, 40, 84 };
        int n = arr.length;
        int k = 3;
        printMax(arr, k, n);
    }
}
 
// This code is contributed by Swetank Modi

Python3

# Python3 program to find k maximum elements 
# of array in original order
 
# Function to print m Maximum elements
def printMax(arr, k, n):
     
    # vector to store the copy of the
    # original array
    brr = arr.copy()
     
    # Sorting the vector in descending
    # order. Please refer below link for
    # details
    brr.sort(reverse = True)
     
    # Traversing through original array and
    # print all those elements that are
    # in first k of sorted vector.
    for i in range(n):
        if (arr[i] in brr[0:k]):
            print(arr[i], end = " ")
 
# Driver code
arr = [ 50, 8, 45, 12, 25, 40, 84 ]
n = len(arr)
k = 3
 
printMax(arr, k, n)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# program to find k maximum  
// elements of array in original order
using System;
using System.Linq;
 
class GFG{
     
// Function to print m Maximum elements
public static void printMax(int[] arr, int k,
                            int n)
{
     
    // Array to store the copy 
    // of the original array
    int[] brr = new int[n];
     
    for(int i = 0; i < n; i++)
        brr[i] = arr[i];
     
    // Sorting the array in 
    // descending order
    Array.Sort(brr);
    Array.Reverse(brr);
     
    int[] crr = new int[k];
     
    for(int i = 0; i < k; i++)
    {
        crr[i] = brr[i];
    }
 
    // Traversing through original array and
    // printing all those elements that are
    // in first k of sorted array.
    // Please refer https://goo.gl/uj5RCD
    // for details of Array.BinarySearch()
    for(int i = 0; i < n; ++i)
    {
        if (crr.Contains(arr[i]))
        {
            Console.Write(arr[i] + " ");
        }
    }
}
 
// Driver code
public static void Main()
{
    int[] arr = { 50, 8, 45, 12, 25, 40, 84 };
    int n = arr.Length;
    int k = 3;
     
    printMax(arr, k, n);
}
}
 
// This code is contributed by Shubhamsingh10

Javascript

<script>
 
// JavaScript program to find k maximum elements 
// of array in original order
 
// Function to print m Maximum elements
function printMax(arr, k, n)
{
    // vector to store the copy of the
    // original array
    var brr = arr.slice();
 
    // Sorting the vector in descending
    // order. Please refer below link for
    // details
    // https://www.zambiatek.com/sort-c-stl/
    brr.sort((a, b) => b - a);
     
    // Traversing through original array and
    // printing all those elements that are
    // in first k of sorted vector.
    // Please refer https://goo.gl/44Rwgt
    // for details of binary_search()
    for (var i = 0; i < n; ++i)
        if (brr.indexOf(arr[i]) < k)
            document.write(arr[i] +" ");
}
 
// Driver code
var arr = [ 50, 8, 45, 12, 25, 40, 84 ];
var n = arr.length;
var k = 3;
printMax(arr, k, n);
 
// This code is contributed by ShubhamSingh10
 
</script>

Output

50 45 84

Time Complexity: O(n Log n) for sorting.
Auxiliary Space: O(n)

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