Sum of the updated array after performing the given operation

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 = 8

Input: 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 array
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
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
int 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 approach
import java.util.*;

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(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 array
def sumArr(arr, n):
    sum = 0
    for i in range(n):
        sum += arr[i]
    return sum

# Function to return the sum
# of the modified array
def 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 code
arr = [40, 25, 12, 10]
n = len(arr)

print(sumModArr(arr, n))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using 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 array
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
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
int 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 approach
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;

        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 array
def 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 code
if __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 approach
using 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)