Check if sum of array can be reduced to zero by repetitively reducing array element by their index value

Given an array arr[] consisting of N integers, the task is to determine if the sum of array elements can be reduced to 0 by performing the following operations any number of times:

• Choose an element A[i] and reduce A[i] by i(1-based indexing), any number of times, possibly 0.
• If the sum can be reduced to 0, print "Yes". Otherwise, print "No".

Examples:

Input: arr[] = {2, 3, 1}
Output: Yes
Explanation:
Select A[2] = 3
Perform given operation 3 times to obtain the following result: 
3 -> (3 -2) -> (3 - 2 - 2) -> (3 - 2 - 2 - 2) 
Sum of the modified array = 2 + (-3) + 1 = 0. 
Therefore, the required answer is 0.

Input: arr[] = {-5, 3}
Output: No

Approach: This problem can be solved based on the following observations:

• If a positive number is repetitively subtracted from another positive number, then eventually, 0 can be obtained. But, repetitively subtracting a positive number from a negative number, 0 can never be obtained as it keeps on decreasing negatively.
• The task to reduce the sum to 0 by subtracting i from A[i].
• Therefore, on choosing an element and reducing the value of the element by i, the sum is being reduced by i. 
   

Let the sum of the array be S.
S = A[1] + A[2] + .... + A[N]
After performing given operations, sum of the array modifies to 
S2 = (A[i] - (i)) + (A[i+1] - (i+1)) .... 
S2 = A[i] + A[i+1].... - (i + i+1....) 
S2 = S - (i + i + 1.....) 
Therefore, after every operation, the original sum is reduced. 

• Therefore, the task reduces to checking if the initial sum of the array is positive or 0. If found to be true, print "Yes". Otherwise, print "No".

Below is the implementation for the above approach:

// C++ Program for the above approach
#include <iostream>
using namespace std;

// Function to check if an array
// sum can be reduced to zero or not
bool isPossible(int arr[], int n)
{
    // Stores the sum of the array
    int S = 0;

    for (int i = 0; i < n; i++) {

        // Update sum of the array
        S = S + arr[i];
    }

    // If the sum is positive
    if (S >= 0) {

        // Array sum can be
        // reduced to 0
        return true;
    }

    // Otherwise
    else {

        // Array sum cannot
        // be reduced to 0
        return false;
    }
}

// Driver Code
int main()
{
    int arr[] = { -5, 3 };

    int n = sizeof(arr) / sizeof(int);

    // Print the answer
    if (isPossible(arr, n)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to check if array sum
    // can be reduced to zero or not
    static boolean isPossible(int[] arr, int n)
    {
        // Stores the sum of the array
        int S = 0;

        for (int i = 0; i < n; i++) {
            S = S + arr[i];
        }

        // If array sum is positive
        if (S >= 0)

            // Array sum can be
            // reduced to 0
            return true;

        // Otherwise
        else

            // Array sum cannot
            // be reduced to 0
            return false;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { -5, 3 };
        int n = arr.length;

        // Function call
        if (isPossible(arr, n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

# Python program for the above approach

# Function to check if an array
# sum can be reduced to zero or not
def isPossible(arr, n):

    # Stores sum of the array
    S = sum(arr)

    # If sum is positive
    if(S >= 0):
        return true

    # If sum is negative
    else:
        return false


# Driver Code
if __name__ == '__main__':

    arr = [-5, 3]
    n = len(arr)

    # Function call
    if (isPossible(arr, n)):
        print("Yes")
    else:
        print("No")

// C# program for
// the above approach
using System;
class GFG{
  
// Function to check if array sum
// can be reduced to zero or not
static bool isPossible(int[] arr, 
                       int n)
{
  // Stores the sum 
  // of the array
  int S = 0;

  for (int i = 0; i < n; i++) 
  {
    S = S + arr[i];
  }

  // If array sum is positive
  if (S >= 0)

    // Array sum can be
    // reduced to 0
    return true;

  // Otherwise
  else

    // Array sum cannot
    // be reduced to 0
    return false;
}

// Driver Code
public static void Main()
{
  int[] arr = {-5, 3};
  int n = arr.Length;

  // Function call
  if (isPossible(arr, n))
    Console.Write("Yes");
  else
    Console.Write("No");
}
}

// This code is contributed by Chitranayal

<script>
// javascript program for the above approach    
// Function to check if array sum
    // can be reduced to zero or not
    function isPossible(arr , n)
    {
    
        // Stores the sum of the array
        var S = 0;

        for (i = 0; i < n; i++) {
            S = S + arr[i];
        }

        // If array sum is positive
        if (S >= 0)

            // Array sum can be
            // reduced to 0
            return true;

        // Otherwise
        else

            // Array sum cannot
            // be reduced to 0
            return false;
    }

    // Driver Code
    
        var arr = [ -5, 3 ];
        var n = arr.length;

        // Function call
        if (isPossible(arr, n))
            document.write("Yes");
        else
            document.write("No");

// This code is contributed by todaysgaurav 
</script>

No

Time Complexity: O(N) 
Auxiliary Space: O(1)