# 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 = 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 + A + …. + 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++

 `// C++ Program for the above approach` `#include ` `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

 `// Java program for teh 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"``);` `    ``}` `}`

## Python3

 `# 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#

 `// 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`

Output:

```No

```

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

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