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. 

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

Javascript

<script>
// Javascript implementation of the approach


// Function to whether the array
// can be made zero or not
function check(arr, N)
{
    // Count for even elements
    let even = 0;

    // Count for odd elements
    let odd = 0;

    // Traverse the array to
    // count the even and odd
    for (let 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)
        document.write("Yes");
    else
        document.write("No");
}

// Driver's Code

let arr = [ 1, 1, 3 ];
let N = arr.length;

check(arr, N);


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

Output:

Yes

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

Auxiliary Space: O(1)