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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

Yes

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

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