# Check if given array can be made 0 with given operations performed any number of times

Given an array arr[] containing N integers, the task is to find whether all the elements of the given array can be made 0 by following operations:

• Increment any element by 2.
• Subtract the minimum element of the array from all elements in the array.
• The above operations can be performed any number times.

If all the elements of the given array can become zero then print Yes else print No.

Examples:

Input: arr[] = {1, 1, 3}
Output: Yes
Explanation:
1st round: Choose the first element in the array and increase it by 2 i.e arr[] = {3, 1, 3}.
Then decrease all the elements by 1(which is minimum in the current array) i.e arr[] = {2, 0, 2}.
2nd round: Choose the second element in the array and increase it by 2 i.e arr[] = {2, 2, 2}.
Then decrease all the elements by 2(which is minimum in the current array) i.e arr[] = {0, 0, 0}.
Therefore, with the given operations performing on the elements of the array, all the elements of the given array can be reduced to 0.

Input: arr[] = {2, 1, 4, 2}
Output: No
Explanation:
We cannot make all the element of the array 0 by performing the given operations.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved with the help of Parity.

• Since, by incrementing the element of the array by 2 in each operation, the parity of the element is not changed i.e., odd remains odd or even remains even.
• And after subtracting each element of the array with the minimum element in the array, the parity of even integers becomes odd and the parity of odd integers becomes even.
• Therefore to make all the elements of the array 0, the parity of all the elements must be same otherwise we can’t make all the elements of the array 0 by the given operations.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to whether the array ` `// can be made zero or not ` `bool` `check(``int` `arr[], ``int` `N) ` `{ ` `    ``// Count for even elements ` `    ``int` `even = 0; ` ` `  `    ``// Count for odd elements ` `    ``int` `odd = 0; ` ` `  `    ``// Traverse the array to ` `    ``// count the even and odd ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``// If arr[i] is odd ` `        ``if` `(arr[i] & 1) { ` `            ``odd++; ` `        ``} ` ` `  `        ``// If arr[i] is even ` `        ``else` `{ ` `            ``even++; ` `        ``} ` `    ``} ` ` `  `    ``// Check if count of even ` `    ``// is zero or count of odd ` `    ``// is zero ` `    ``if` `(even == N || odd == N) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `} ` ` `  `// Driver's Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 1, 3 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``check(arr, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to whether the array ` `// can be made zero or not ` `static` `void` `check(``int` `arr[], ``int` `N) ` `{ ` `    ``// Count for even elements ` `    ``int` `even = ``0``; ` `  `  `    ``// Count for odd elements ` `    ``int` `odd = ``0``; ` `  `  `    ``// Traverse the array to ` `    ``// count the even and odd ` `    ``for` `(``int` `i = ``0``; i < N; i++) { ` `  `  `        ``// If arr[i] is odd ` `        ``if` `(arr[i] % ``2` `== ``1``) { ` `            ``odd++; ` `        ``} ` `  `  `        ``// If arr[i] is even ` `        ``else` `{ ` `            ``even++; ` `        ``} ` `    ``} ` `  `  `    ``// Check if count of even ` `    ``// is zero or count of odd ` `    ``// is zero ` `    ``if` `(even == N || odd == N) ` `        ``System.out.print(``"Yes"``); ` `    ``else` `        ``System.out.print(``"No"``); ` `} ` `  `  `// Driver's Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``1``, ``1``, ``3` `}; ` `    ``int` `N = arr.length; ` `  `  `    ``check(arr, N);  ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to whether the array ` `# can be made zero or not ` `def` `check(arr, N): ` `     `  `    ``# Count for even elements ` `    ``even ``=` `0``; ` ` `  `    ``# Count for odd elements ` `    ``odd ``=` `0``; ` ` `  `    ``# Traverse the array to ` `    ``# count the even and odd ` `    ``for` `i ``in` `range``(N): ` ` `  `        ``# If arr[i] is odd ` `        ``if` `(arr[i] ``%` `2` `=``=` `1``): ` `            ``odd ``+``=` `1``; ` `     `  `        ``# If arr[i] is even ` `        ``else``: ` `            ``even ``+``=` `1``; ` ` `  `    ``# Check if count of even ` `    ``# is zero or count of odd ` `    ``# is zero ` `    ``if` `(even ``=``=` `N ``or` `odd ``=``=` `N): ` `        ``print``(``"Yes"``); ` `    ``else``: ` `        ``print``(``"No"``); ` ` `  `# Driver's Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[ ``1``, ``1``, ``3``]; ` `    ``N ``=` `len``(arr); ` ` `  `    ``check(arr, N); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG{ ` `   `  `// Function to whether the array ` `// can be made zero or not ` `static` `void` `check(``int` `[]arr, ``int` `N) ` `{ ` `    ``// Count for even elements ` `    ``int` `even = 0; ` `   `  `    ``// Count for odd elements ` `    ``int` `odd = 0; ` `   `  `    ``// Traverse the array to ` `    ``// count the even and odd ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `   `  `        ``// If arr[i] is odd ` `        ``if` `(arr[i] % 2 == 1) { ` `            ``odd++; ` `        ``} ` `   `  `        ``// If arr[i] is even ` `        ``else` `{ ` `            ``even++; ` `        ``} ` `    ``} ` `   `  `    ``// Check if count of even ` `    ``// is zero or count of odd ` `    ``// is zero ` `    ``if` `(even == N || odd == N) ` `        ``Console.Write(``"Yes"``); ` `    ``else` `        ``Console.Write(``"No"``); ` `} ` `   `  `// Driver's Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 1, 1, 3 }; ` `    ``int` `N = arr.Length; ` `   `  `    ``check(arr, N);  ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```Yes
```

Time Complexity: O(N), where N is the length of the given array.

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Improved By : Rajput-Ji, 29AjayKumar

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