Replace every element of array with sum of elements on its right side

Last Updated : 28 Mar, 2023

Given an array arr[], the task is to replace every element of the array with the sum of elements on its right side.
Examples:

Input: arr[] = {1, 2, 5, 2, 2, 5}
Output: 16 14 9 7 5 0

Input: arr[] = {5, 1, 3, 2, 4}
Output: 10 9 6 4 0

Naive Approach: A simple approach is to run two loops, Outer loop to fix each element one by one and inner loop to calculate sum of elements on right side of fixed element.

C++

// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)
// C++ program to Replace every element of array
// with sum of elements on its right side
#include<bits/stdc++.h>
using namespace std;
 
// Function to replace every element
// of the array to the sum of elements
// on the right side of the array
void replaceElement(int arr[], int n){
    for(int i = 0; i<n; i++){
        int sum = 0;
        for(int j = i+1; j<n; j++){
            sum += arr[j];
        }
        arr[i] = sum;
    }
    
    for(int i = 0; i<n; i++){
        cout<<arr[i]<<" ";
    }
}
 
// Driver code to test above function
int main(){
    int arr[] = { 1, 2, 5, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    replaceElement(arr, n);
}

Java

public class GFG {
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 5, 2, 2, 5 };
        int n = arr.length;

        replaceElement(arr, n);
    }

    /**
     * Function to replace every element of the array to the
     * sum of elements on the right side of the array
     * @param arr the input array of integers
     * @param n the size of the array
     */
    public static void replaceElement(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i + 1; j < n; j++) {
                sum += arr[j];
            }
            arr[i] = sum;
        }

        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
    }
}

Python3

# Function to replace every element
# of the array to the sum of elements
# on the right side of the array
def replaceElement(arr, n):
    for i in range(n):
        sum = 0
        for j in range(i+1, n):
            sum += arr[j]
        arr[i] = sum

    for i in range(n):
        print(arr[i], end=" ")

# Driver code to test above function
arr = [1, 2, 5, 2, 2, 5]
n = len(arr)

replaceElement(arr, n)

using System;

namespace GFG {
class Program {
    static void Main(string[] args)
    {
        int[] arr = { 1, 2, 5, 2, 2, 5 };
        int n = arr.Length;

        ReplaceElement(arr, n);
    }

    /**
     * Function to replace every element of the array to the
     * sum of elements on the right side of the array
     * @param arr the input array of integers
     * @param n the size of the array
     */
    static void ReplaceElement(int[] arr, int n)
    {
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i + 1; j < n; j++) {
                sum += arr[j];
            }
            arr[i] = sum;
        }

        for (int i = 0; i < n; i++) {
            Console.Write(arr[i] + " ");
        }
    }
}
}
// This code is contributed by Prajwal Kandekar

JavaScript

// Function to replace every element of the array to the sum of elements on the right side of the array
function replaceElement(arr) {
    for (let i = 0; i < arr.length; i++) {
        let sum = 0;
        for (let j = i + 1; j < arr.length; j++) {
            sum += arr[j];
        }
        arr[i] = sum;
    }
    console.log(arr.join(" "));
}

// Driver code to test above function
let arr = [1, 2, 5, 2, 2, 5];
replaceElement(arr);

Output

16 14 9 7 5 0

Time Complexity: O(N^2)

Auxiliary Space : O(1)
Efficient Approach: The idea is to compute the total sum of the array and then update the current element as the sum - arr[i] and in each step update the sum to the current element.

for i in arr:
  arr[i] = sum - arr[i]
  sum = arr[i]

Below is the implementation of the above approach:

C++

// C++ program to Replace every
// element of array with sum of
// elements on its right side
#include<bits/stdc++.h>
using namespace std;

// Function to replace every element
// of the array to the sum of elements
// on the right side of the array
void replaceElement(int arr[], int n)
{
    int sum = 0;

    // Calculate sum of all
    // elements of array
    for(int i = 0; i < n; i++)
       sum += arr[i];

    // Traverse the array
    for(int i = 0; i < n; i++)
    {
        
       // Replace current element
       // of array with sum-arr[i]
       arr[i] = sum - arr[i];
       
       // Update sum with arr[i]
       sum = arr[i];
    }
    
    // Print modified array
    for(int i = 0; i < n; i++)
       cout << arr[i] << " ";
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 5, 2, 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    
    replaceElement(arr, n);
}

// This code is contributed by Surendra_Gangwar

Java

// Java program to Replace every
// element of array with sum of
// elements on its right side

import java.util.*;
class GFG {

    // Function to replace every element
    // of the array to the sum of elements
    // on the right side of the array
    static void replaceElement(int[] arr, int n)
    {
        int sum = 0;

        // Calculate sum of all
        // elements of array
        for (int i = 0; i < n; i++)
            sum += arr[i];

        // Traverse the array
        for (int i = 0; i < n; i++) {

            // Replace current element
            // of array with sum-arr[i]
            arr[i] = sum - arr[i];

            // Update sum with arr[i]
            sum = arr[i];
        }

        // Print modified array
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 5, 2, 2, 5 };
        int n = arr.length;
        replaceElement(arr, n);
    }
}

Python3

# Python3 program to replace every
# element of array with sum of
# elements on its right side

# Function to replace every element
# of the array to the sum of elements
# on the right side of the array
def replaceElement(arr, n):
    
    sum = 0;

    # Calculate sum of all
    # elements of array
    for i in range(0, n):
        sum += arr[i];
        
    # Traverse the array
    for i in range(0, n):
        
        # Replace current element
        # of array with sum-arr[i]
        arr[i] = sum - arr[i];
        
        # Update sum with arr[i]
        sum = arr[i];
    
    # Print modified array
    for i in range(0, n):
        print(arr[i], end = " ");

# Driver Code 
if __name__ == "__main__": 
    
    arr = [ 1, 2, 5, 2, 2, 5 ]
    n = len(arr)
    
    replaceElement(arr, n)
    
# This code is contributed by Ritik Bansal

// C# program to Replace every
// element of array with sum of
// elements on its right side
using System;
class GFG{

// Function to replace every element
// of the array to the sum of elements
// on the right side of the array
static void replaceElement(int[] arr, int n)
{
    int sum = 0;

    // Calculate sum of all
    // elements of array
    for (int i = 0; i < n; i++)
        sum += arr[i];

    // Traverse the array
    for (int i = 0; i < n; i++) 
    {

        // Replace current element
        // of array with sum-arr[i]
        arr[i] = sum - arr[i];

        // Update sum with arr[i]
        sum = arr[i];
    }

    // Print modified array
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}

// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 5, 2, 2, 5 };
    int n = arr.Length;
    replaceElement(arr, n);
}
}

// This code is contributed by Nidhi_biet

JavaScript

<script>

// JavaScript program to Replace every
// element of array with sum of
// elements on its right side

// Function to replace every element
// of the array to the sum of elements
// on the right side of the array
function replaceElement(arr, n)
{
    let sum = 0;

    // Calculate sum of all
    // elements of array
    for(let i = 0; i < n; i++)
    sum += arr[i];

    // Traverse the array
    for(let i = 0; i < n; i++)
    {
        
    // Replace current element
    // of array with sum-arr[i]
    arr[i] = sum - arr[i];
        
    // Update sum with arr[i]
    sum = arr[i];
    }
    
    // Print modified array
    for(let i = 0; i < n; i++)
    document.write(arr[i] + " ");
}

// Driver code

    let arr = [ 1, 2, 5, 2, 2, 5 ];
    let n = arr.length;
    
    replaceElement(arr, n);

//This code is contributed by Mayank Tyagi

</script>

Output:

16 14 9 7 5 0

Time Complexity: O(N)

Auxiliary Space: O(1)

Sum of Bitwise OR of every array element paired with all other array elements

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Replace every element of array with sum of elements on its right side

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