Replace every element of array with sum of elements on its right side
Last Updated :
15 Jul, 2025
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);
}
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] + " ");
}
}
}
# 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
// 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);
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 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 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(1)
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