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 0Input: 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 arrayvoid 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 functionint 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 arraydef 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 functionarr = [1, 2, 5, 2, 2, 5]n = len(arr)replaceElement(arr, n) |
C#
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 arrayfunction 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 functionlet 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]](https://cdn.geeksforgeeks.org/static/gallery/images/logo.png "Rendered by QuickLaTeX.com"){kind=logo}
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 arrayvoid 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 codeint 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 sideimport 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 arraydef 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#
// C# program to Replace every// element of array with sum of// elements on its right sideusing System;class GFG{// Function to replace every element// of the array to the sum of elements// on the right side of the arraystatic 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 codepublic 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 arrayfunction 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)
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