Find Kth largest element from right of every element in the array

Given an array arr[] of size N and an integer K. The task is to find the Kth largest element from the right of every element in the array. If there are not enough elements to the right then print the same element.

Examples:

Input: N = 6, K = 3, arr[] = {4, 5, 3, 6, 7, 2}
Output: 5 3 2 6 7 2
Explanation: The elements right to 4 are {5, 3, 6, 7, 2}. 
So 3rd largest element to the right of 4 is 5. 
Similarly, repeat the process for the rest of the elements.
And 7 and 2 does not have sufficient element to the right.
So, they are kept as it is.

Input: N = 5, K = 2, arr[] = {-4, 7, 5, 3, 0}
Output: 5 3 0 3 0

Naive Approach: The naive approach is to sort every subarray to the right of every element and check whether the Kth largest element exists or not. If it exists, print the Kth largest element, else print the same element.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the Kth
// largest element to the right
// of every element
int getKLargest(int* arr, int r,
                int l, int& K)
{
    // Elements to the right
    // of current element
    vector<int> v(arr, arr + l + 1);

    // There are greater than K elements
    // to the right
    if (l - r >= K) {

        // Sort the vector
        sort(v.begin() + r + 1, v.end());
        return v[l - K + 1];
    }
    else
        return v[r];
}

// Driver Code
int main()
{
    int arr[] = { -4, 7, 5, 3, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;

    for (int i = 0; i < N; i++)
        cout << getKLargest(arr, i, N - 1, K)
             << " ";

    return 0;
}

Java

// Java code for the above approach
import java.util.*;

class GFG{

  // Function to sort the elements of the array
  // from index a to index b
  static void sort(int[] arr, int N, int a, int b)
  {

    // Variables to store start and end of the index
    // range
    int l = Math.min(a, b);
    int r = Math.max(a, b);

    // Temporary array
    int[] temp = new int[r - l + 1];
    int j = 0;
    for (int i = l; i <= r; i++) {
      temp[j] = arr[i];
      j++;
    }

    // Sort the temporary array
    Arrays.sort(temp);

    // Modifying original array with temporary array
    // elements
    j = 0;
    for (int i = l; i <= r; i++) {
      arr[i] = temp[j];
      j++;
    }
  }
  // Function to find the Kth
  // largest element to the right
  // of every element
  static int getKLargest(int[] arr, int r, int l, int K)
  {
    // Elements to the right
    // of current element
    int n = arr.length;
    int[] v = new int[l + 1];
    for (int i = 0; i < l + 1; i++) {
      v[i] = arr[i];
    }

    // There are greater than K elements
    // to the right
    if (l - r >= K) {

      // Sort the vector
      sort(v, n, r + 1, n - 1);
      return v[l - K + 1];
    }
    else
      return v[r];
  }

  // Driver Code
  public static void main(String[] args)
  {
    int arr[] = { -4, 7, 5, 3, 0 };
    int N = arr.length;
    int K = 2;

    for (int i = 0; i < N; i++)
      System.out.print(getKLargest(arr, i, N - 1, K)
                       + " ");
  }
}

// This code is contributed by code_hunt.

Python3

# Python code for the above approach

# Function to find the Kth
# largest element to the right
# of every element
def getKLargest(arr, r, l, K):

    # Elements to the right
    # of current element
    v = arr[0: l + 1]

    # There are greater than K elements
    # to the right
    if (l - r >= K):

        # Sort the vector
        temp1 = v[0: r + 1]

        temp = v[r + 1:]
        temp.sort()
        v = temp1 + temp
        return v[l - K + 1]
    else:
        return v[r]

# Driver Code
arr = [-4, 7, 5, 3, 0]
N = len(arr)
K = 2

for i in range(N):
    print(getKLargest(arr, i, N - 1, K), end=" ")

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG
{

  // Function to sort the elements of the array
  // from index a to index b
  static void sort(int[] arr, int N, int a, int b)
  {

    // Variables to store start and end of the index
    // range
    int l = Math.Min(a, b);
    int r = Math.Max(a, b);

    // Temporary array
    int[] temp = new int[r - l + 1];
    int j = 0;
    for (int i = l; i <= r; i++) {
      temp[j] = arr[i];
      j++;
    }

    // Sort the temporary array
    Array.Sort(temp);

    // Modifying original array with temporary array
    // elements
    j = 0;
    for (int i = l; i <= r; i++) {
      arr[i] = temp[j];
      j++;
    }
  }
  // Function to find the Kth
  // largest element to the right
  // of every element
  static int getKLargest(int[] arr, int r, int l, int K)
  {
    // Elements to the right
    // of current element
    int n = arr.Length;
    int[] v = new int[l + 1];
    for (int i = 0; i < l + 1; i++) {
      v[i] = arr[i];
    }

    // There are greater than K elements
    // to the right
    if (l - r >= K) {

      // Sort the vector
      sort(v, n, r + 1, n - 1);
      return v[l - K + 1];
    }
    else
      return v[r];
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { -4, 7, 5, 3, 0 };
    int N = arr.Length;
    int K = 2;

    for (int i = 0; i < N; i++)
      Console.Write(getKLargest(arr, i, N - 1, K)
                    + " ");
  }
}
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
        // JavaScript code for the above approach

        // Function to find the Kth
        // largest element to the right
        // of every element
        function getKLargest(arr, r,
            l, K)
        {

            // Elements to the right
            // of current element
            let v = arr.slice(0, l + 1);

            // There are greater than K elements
            // to the right
            if (l - r >= K) {

                // Sort the vector
                let temp1 = v.slice(0, r + 1);

                let temp = v.slice(r + 1);
                temp.sort(function (a, b) { return a - b })
                v = temp1.concat(temp)
                return v[l - K + 1];
            }
            else
                return v[r];
        }

        // Driver Code
        let arr = [-4, 7, 5, 3, 0];
        let N = arr.length;
        let K = 2;

        for (let i = 0; i < N; i++)
            document.write(getKLargest(arr, i, N - 1, K) + " ")

         // This code is contributed by Potta Lokesh
    </script>

Output

5 3 0 3 0 

Time Complexity: O(N² * logN)
Auxiliary Space: O(N) 

Approach based on Sets: Another approach is to use sets. Though both the approaches have same the time complexity, sets have an inbuilt sorting feature, it doesn’t require explicit sorting.

Below is the implementation of the above approach.

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the kth largest
// element to the right
int getKLargest(vector<int>& arr, int r,
                int l, int& k)
{
    set<int> s(arr.begin() + r + 1,
               arr.end());
    if (l - r >= k) {
        set<int>::iterator it = s.end();
        advance(it, -k);
        return *it;
    }
    else
        return arr[r];
}

// Driver code
int main()
{
    int arr[] = { -4, 7, 5, 3, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int i, K = 2;

    vector<int> a(arr, arr + N);

    for (i = 0; i < N; i++)
        cout << getKLargest(a, i, N - 1, K)
             << " ";

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the kth largest
// element to the right
static int getKLargest(List<Integer> arr, int r,
                int l, int k)
{
    HashSet<Integer> s = new HashSet<>(arr.subList(r+1,arr.size()));
    List<Integer> a = new ArrayList<>(s);
    if (l - r >= k) {
        return a.get(a.size()-k);
    }
    else
        return arr.get(r);
}

// Driver code
public static void main(String[] args)
{
    Integer arr[] = { -4, 7, 5, 3, 0 };
    int N = arr.length;
    int i=0;
    int K = 2;

    List<Integer> a = Arrays.asList(arr);

    for (i = 0; i < N; i++)
        System.out.print(getKLargest(a, i, N - 1, K)
            + " ");
}
}

// This code is contributed by shikhasingrajput

Python3

#  Python3 program for the above approach

# def to find the kth largest
# element to the right
def getKLargest( arr , r , l , k):
   s = set()
   for i in range(r+1,len(arr)):
      s.add(arr[i])
   a = []
   for p in s:
      a.append(p)

   if(l - r >= k):
      return a[len(a)- k]
   else:
      return arr[r]

    #  Driver code

arr = [ -4, 7, 5, 3, 0 ]
N = len(arr)
i = 0
K = 2

a = arr

for i in range(N):
    print(getKLargest(a, i, N - 1, K) ,end=" ")

#  This code is contributed by shinjanpatra

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG{

  // Function to find the kth largest
  // element to the right
  static int getKLargest(List<int> arr, int r,
                         int l, int k)
  {

    HashSet<int> s = new HashSet<int>(arr.GetRange(r+1,arr.Count-(r+1)));
    List<int> a = new List<int>(s);
    a.Reverse();
    if (l - r >= k) {
      return a[a.Count-k];
    }
    else
      return arr[r];
  }

  // Driver code
  public static void Main(String[] args)
  {
    int []arr = { -4, 7, 5, 3, 0 };
    int N = arr.Length;
    int i=0;
    int K = 2;

    List<int> a = new List<int>(arr);

    for (i = 0; i < N; i++)
      Console.Write(getKLargest(a, i, N - 1, K)
                    + " ");
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// javascript program for the above approach

    // Function to find the kth largest
    // element to the right
    function getKLargest( arr , r , l , k) {
        var s = new Set();
        for(let  i = r+1;i<arr.length;i++)
        s.add(arr[i]);
      //  <>(arr.subList(r + 1, arr.size()));
        var a = [];
        for(let p of s)
        a.push(p);
        a.reverse();
        if (l - r >= k) {
            return a[a.length - k];
        } else
            return arr[r];
    }

    // Driver code

        var arr = [ -4, 7, 5, 3, 0 ];
        var N = arr.length;
        var i = 0;
        var K = 2;

        var a = (arr);

        for (var i = 0; i < N; i++)
            document.write(getKLargest(a, i, N - 1, K) + " ");

// This code contributed by Rajput-Ji
</script>

Output

5 3 0 3 0 

Time Complexity: O(N² * logN)
Auxiliary Space: O(N)

Efficient Approach: The idea to efficiently solve the problem is:

Insert each element of the array in a vector and sort them in decreasing order. Then delete each element and print the Kth largest from the sorted vector.

Follow the steps mentioned below to implement the idea:

• First, insert all the elements in a vector.
• Sort the vector in decreasing order.
• Iterate through the array from i = 0 to N:
  • Delete arr[i] from the vector.
  • If there are at least K elements left, print the Kth largest from the vector.
  • Otherwise, print the same element.

Below is the implementation of the above approach.

C++

// C++ code for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to print the Kth
// largest element to the right
// of every element
int getKLargest(int arr[], int n, int k)
{
    // Storing elements in a vector
    vector<int> v(arr, arr + n );

    // Sort the vector in decreasing order
    sort(v.begin(),v.end(),greater<int>());

    // Iterating the for loop
    // to find kth largest element
    // for every element in the array
    for(int i=0;i<n;i++)
    {
        // Erase the current element from the vector
        v.erase(remove(v.begin(), v.end(), arr[i]), v.end());

        // If kth largest element is possible
        // then print it.
        if(v.size()>=k)
        {
            cout<<v[k-1]<<" ";
        }

        // If kth largest is not possible then
        // print the element in the array.
        else
        {
            cout<<arr[i]<<" ";
        }
    }
}

// Driver Code
int main()
{
    int arr[] = { -4, 7, 5, 3, 0 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 2;

    // Function call
    getKLargest(arr, N , K);

    return 0;
}

// This code is contributed by Pushpesh raj.

Java

// Java code for the above approach
import java.util.*;

class GFG {

  // Function to print the Kth
  // largest element to the right
  // of every element
  static void getKLargest(int arr[],
                          int n, int k)
  {
    // Storing elements in a vector
    Vector<Integer> v
      = new Vector<Integer>();
    for (int i = 0; i < n; i++)
      v.add(arr[i]);

    // Sort the vector in decreasing order
    Collections.sort(v, Collections.reverseOrder());

    // Iterating the for loop
    // to find kth largest element
    // for every element in the array
    for (int i = 0; i < n; i++)
    {
      // Erase the current element
      // from the vector
      v.remove(new Integer(arr[i]));

      // If kth largest element is possible
      // then print it.
      if (v.size() >= k)
        System.out.print(v.get(k - 1) + " ");

      // If kth largest is not
      // possible then print the
      // element in the array
      else
        System.out.print(arr[i] + " ");
    }
  }

  // Driver Code
  public static void main(String args[])
  {
    int arr[] = { -4, 7, 5, 3, 0 };
    int N = arr.length;
    int K = 2;

    // Function call
    getKLargest(arr, N , K);
  }
}

// This code is contributed by adityamaharshi21

Python3

# Python3 program for the above approach:

## Function to print the Kth
## largest element to the right
## of every element
def getKLargest(arr, n, k):

    ## Storing elements in a vector
    v = []
    for i in arr:
        v.append(i)

    ## Sort the vector in decreasing order
    v.sort()
    v.reverse()

    ## Iterating the for loop
    ## to find kth largest element
    ## for every element in the array
    for i in range(n):

        ## Erase the current element from the vector
        v.remove(arr[i])

        ## If kth largest element is possible
        ## then print it.
        if(len(v) >= k):
            print(v[k-1], end = ' ')

        ## If kth largest is not possible then
        ## print the element in the array.
        else:
            print(arr[i], end = ' ')

## Driver code
if __name__=='__main__':

    arr = [-4, 7, 5, 3, 0]
    N = len(arr)
    K = 2

    ## Function call
    getKLargest(arr, N , K)

    # This code is contributed by subhamgoyal2014.

C#

// C# code for the above approach
using System;
using System.Linq;

// Function to print the Kth
// largest element to the right
// of every element
public class GFG {
  public static void getKLargest(int[] arr, int n, int k)
  {

    // Storing elements in a List
    var v = arr.ToList();

    // Sort the List in decreasing order
    v.Sort((a, b) = > b - a);

    // Iterating the for loop
    // to find kth largest element
    // for every element in the array
    for (int i = 0; i < n; i++) {
      // Erase the current element from the List
      v.Remove(arr[i]);

      // If kth largest element is possible
      // then print it.
      if (v.Count >= k) {
        Console.Write(v[k - 1] + " ");
      }

      // If kth largest is not possible then
      // print the element in the array.
      else {
        Console.Write(arr[i] + " ");
      }
    }
  }

  // Driver Code
  public static void Main()
  {
    int[] arr = { -4, 7, 5, 3, 0 };
    int N = arr.Length;
    int K = 2;

    // Function call
    getKLargest(arr, N, K);
  }
}

Javascript

<script>

// JavaScript code for the above approach


// Function to print the Kth
// largest element to the right
// of every element
function getKLargest(arr, n, k)
{
    // Storing elements in a vector
    let v = arr.slice();

    // Sort the vector in decreasing order
    v.sort((a,b)=>b-a);

    // Iterating the for loop
    // to find kth largest element
    // for every element in the array
    for(let i=0;i<n;i++)
    {
        // Erase the current element from the vector
        v.splice(v.indexOf(arr[i]),1);

        // If kth largest element is possible
        // then print it.
        if(v.length>=k)
        {
            document.write(v[k-1]," ");
        }

        // If kth largest is not possible then
        // print the element in the array.
        else
        {
            document.write(arr[i]," ");
        }
    }
}

// Driver Code

let arr = [ -4, 7, 5, 3, 0 ];
let N = arr.length;
let K = 2;

// Function call
getKLargest(arr, N , K);

// This code is contributed by shinjanpatra

</script>

Output

5 3 0 3 0 

Time Complexity: O(N²)
Auxiliary Space: O(N)