检查是否可以将给定的数组执行任意次数，使给定的数组为0

给定一个包含N个整数的数组arr [] ，任务是通过以下操作查找给定数组的所有元素是否可以设为0：

• 将任何元素加2。
• 从数组中的所有元素中减去数组中的最小元素。
• 以上操作可以执行任意次。

如果给定数组的所有元素都可以变为零，则打印“是”，否则打印“否” 。

例子：

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.

方法：可以通过奇偶校验解决问题。

• 由于通过在每次操作中将数组的元素加2，该元素的奇偶校验不变，即，奇数保持为奇数，甚至偶数保持为。
• 在用数组中的最小元素减去数组的每个元素之后，偶数整数的奇偶校验变为奇数，奇数整数的奇偶校验变为偶数。
• 因此，要使数组0的所有元素，所有元素的奇偶性必须相同，否则我们不能通过给定的操作使数组0的所有元素。

下面是上述方法的实现：

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[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

输出：

Yes

时间复杂度： O(N) ，其中N是给定数组的长度。