Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 28 May, 2021

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.
 

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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :