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:YesExplanation:

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++

`// 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

`// 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` |

## Javascript

`<script>` `// javascript program for teh 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>` |

**Output:**

No

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

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**