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:YesExplanation:

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:NoExplanation:

We cannot make all the element of the array 0 by performing the given operations.

**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 <bits/stdc++.h>` `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[0]);` ` ` ` ` `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.

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****.**