Given an array arr[] of N elements, the task is to update all the array elements such that an element arr[i] is updated as arr[i] = arr[i] – X where X = arr[i + 1] + arr[i + 2] + … + arr[N – 1] and finally print the sum of the updated array.
Examples:
Input: arr[] = {40, 25, 12, 10}
Output: 8
The updated array will be {-7, 3, 2, 10}.
-7 + 3 + 2 + 10 = 8Input: arr[] = {50, 30, 10, 2, 0}
Output: 36
Approach: A simple solution is for every possible value of i, update arr[i] = arr[i] – sum(arr[i+1…N-1]).
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Utility function to return// the sum of the arrayint sumArr(int arr[], int n){ int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum;}// Function to return the sum// of the modified arrayint sumModArr(int arr[], int n){ for (int i = 0; i < n - 1; i++) { // Find the sum of the subarray // arr[i+1...n-1] int subSum = 0; for (int j = i + 1; j < n; j++) { subSum += arr[j]; } // Subtract the subarray sum arr[i] -= subSum; } // Return the sum of // the modified array return sumArr(arr, n);}// Driver codeint main(){ int arr[] = { 40, 25, 12, 10 }; int n = sizeof(arr) / sizeof(arr[0]); cout << sumModArr(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{// Utility function to return// the sum of the arraystatic int sumArr(int arr[], int n){ int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum;}// Function to return the sum// of the modified arraystatic int sumModArr(int arr[], int n){ for (int i = 0; i < n - 1; i++) { // Find the sum of the subarray // arr[i+1...n-1] int subSum = 0; for (int j = i + 1; j < n; j++) { subSum += arr[j]; } // Subtract the subarray sum arr[i] -= subSum; } // Return the sum of // the modified array return sumArr(arr, n);}// Driver codepublic static void main(String []args){ int arr[] = { 40, 25, 12, 10 }; int n = arr.length; System.out.println(sumModArr(arr, n));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach# Utility function to return# the sum of the arraydef sumArr(arr, n): sum = 0 for i in range(n): sum += arr[i] return sum# Function to return the sum# of the modified arraydef sumModArr(arr, n): for i in range(n - 1): # Find the sum of the subarray # arr[i+1...n-1] subSum = 0 for j in range(i + 1, n): subSum += arr[j] # Subtract the subarray sum arr[i] -= subSum # Return the sum of # the modified array return sumArr(arr, n)# Driver codearr = [40, 25, 12, 10]n = len(arr)print(sumModArr(arr, n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{ // Utility function to return // the sum of the array static int sumArr(int []arr, int n) { int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum; } // Function to return the sum // of the modified array static int sumModArr(int []arr, int n) { for (int i = 0; i < n - 1; i++) { // Find the sum of the subarray // arr[i+1...n-1] int subSum = 0; for (int j = i + 1; j < n; j++) { subSum += arr[j]; } // Subtract the subarray sum arr[i] -= subSum; } // Return the sum of // the modified array return sumArr(arr, n); } // Driver code public static void Main() { int []arr = { 40, 25, 12, 10 }; int n = arr.Length; Console.WriteLine(sumModArr(arr, n)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// javascript implementation of the approach // Utility function to return // the sum of the array function sumArr(arr , n) { var sum = 0; for (i = 0; i < n; i++) sum += arr[i]; return sum; } // Function to return the sum // of the modified array function sumModArr(arr , n) { for (i = 0; i < n - 1; i++) { // Find the sum of the subarray // arr[i+1...n-1] var subSum = 0; for (j = i + 1; j < n; j++) { subSum += arr[j]; } // Subtract the subarray sum arr[i] -= subSum; } // Return the sum of // the modified array return sumArr(arr, n); } // Driver code var arr = [ 40, 25, 12, 10 ]; var n = arr.length; document.write(sumModArr(arr, n));// This code is contributed by todaysgaurav</script> |
Output:
8
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient approach: An efficient solution is to traverse the array from the end so that the sum of the subarray till now i.e. sum(arr[i+1…n-1]) can be used to calculate the sum of the current subarray arr[i…n-1] i.e. sum(arr[i…n-1]) = arr[i] + sum(arr[i+1…n-1]).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Utility function to return// the sum of the arrayint sumArr(int arr[], int n){ int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum;}// Function to return the sum// of the modified arrayint sumModArr(int arr[], int n){ int subSum = arr[n - 1]; for (int i = n - 2; i >= 0; i--) { int curr = arr[i]; // Subtract the subarray sum arr[i] -= subSum; // Sum of subarray arr[i...n-1] subSum += curr; } // Return the sum of // the modified array return sumArr(arr, n);}// Driver codeint main(){ int arr[] = { 40, 25, 12, 10 }; int n = sizeof(arr) / sizeof(arr[0]); cout << sumModArr(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG { // Utility function to return // the sum of the array static int sumArr(int arr[], int n) { int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum; } // Function to return the sum // of the modified array static int sumModArr(int arr[], int n) { int subSum = arr[n - 1]; for (int i = n - 2; i >= 0; i--) { int curr = arr[i]; // Subtract the subarray sum arr[i] -= subSum; // Sum of subarray arr[i...n-1] subSum += curr; } // Return the sum of // the modified array return sumArr(arr, n); } // Driver code public static void main (String[] args) { int []arr = { 40, 25, 12, 10 }; int n = arr.length; System.out.println(sumModArr(arr, n)); } }// This code is contributed by kanugargng |
Python3
# Python3 implementation of the approach# Function to return the sum# of the modified arraydef sumModArr(arr, n): subSum = arr[n - 1]; for i in range(n - 2, -1, -1): curr = arr[i]; # Subtract the subarray sum arr[i] -= subSum; # Sum of subarray arr[i...n-1] subSum += curr; # Return the sum of # the modified array return sum(arr)# Driver codeif __name__ == "__main__": arr = [40, 25, 12, 10 ]; n = len(arr); print(sumModArr(arr, n));# This code is contributed by Shushant Kumar |
C#
// C# implementation of the approachusing System; class GFG { // Utility function to return // the sum of the array static int sumArr(int []arr, int n) { int sum = 0; for (int i = 0; i < n; i++) sum += arr[i]; return sum; } // Function to return the sum // of the modified array static int sumModArr(int []arr, int n) { int subSum = arr[n - 1]; for (int i = n - 2; i >= 0; i--) { int curr = arr[i]; // Subtract the subarray sum arr[i] -= subSum; // Sum of subarray arr[i...n-1] subSum += curr; } // Return the sum of // the modified array return sumArr(arr, n); } // Driver code public static void Main (String[] args) { int []arr = { 40, 25, 12, 10 }; int n = arr.Length; Console.WriteLine(sumModArr(arr, n)); } }// This code is contributed by 29AjayKumar |
Javascript
<script> // JavaScript implementation of the approach // Utility function to return // the sum of the array function sumArr(arr, n) { var sum = 0; for (var i = 0; i < n; i++) sum += arr[i]; return sum; } // Function to return the sum // of the modified array function sumModArr(arr, n) { var subSum = arr[n - 1]; for (var i = n - 2; i >= 0; i--) { var curr = arr[i]; // Subtract the subarray sum arr[i] -= subSum; // Sum of subarray arr[i...n-1] subSum += curr; } // Return the sum of // the modified array return sumArr(arr, n); } // Driver code var arr = [40, 25, 12, 10]; var n = arr.length; document.write(sumModArr(arr, n)); // This code is contributed by rdtank. </script> |
Output:
8
Time Complexity: O(N)
Auxiliary Space: O(1)
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