Check if all the elements can be made of same parity by inverting adjacent elements

Given a binary matrix. In a single operation, you are allowed to choose two adjacent elements and invert their parity. The operation can be performed any number of times. Write a program to check if all the elements of the array can be converted into a single parity.

Examples:

Input: a[] = {1, 0, 1, 1, 0, 1}
Output: Yes
Invert 2nd and 3rd elements to get {1, 1, 0, 1, 0, 1}
Invert 3rd and 4th elements to get {1, 1, 1, 0, 0, 1}
Invert 4th and 5th elements to get {1, 1, 1, 1, 1, 1}



Input: a[] = {1, 1, 1, 0, 0, 0}
Output: No

Approach: Since only adjacent elements are needed to be flipped, hence the count of parities will give the answer to the question. Only even number of elements are flipped at a time, so if both the parity’s count is odd then it is not possible to make all the parity same else it is possible.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if parity
// can be made same or not
bool flipsPossible(int a[], int n)
{
  
    int count_odd = 0, count_even = 0;
  
    // Iterate and count the parity
    for (int i = 0; i < n; i++) {
  
        // Odd
        if (a[i] & 1)
            count_odd++;
  
        // Even
        else
            count_even++;
    }
  
    // Condition check
    if (count_odd % 2 && count_even % 2)
        return false;
  
    else
        return true;
}
  
// Drivers code
int main()
{
    int a[] = { 1, 0, 1, 1, 0, 1 };
    int n = sizeof(a) / sizeof(a[0]);
  
    if (flipsPossible(a, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach 
public class GFG 
{
      
    // Function to check if parity 
    // can be made same or not 
    static boolean flipsPossible(int []a, int n) 
    
      
        int count_odd = 0, count_even = 0
      
        // Iterate and count the parity 
        for (int i = 0; i < n; i++) 
        
      
            // Odd 
            if ((a[i] & 1) == 1
                count_odd++; 
      
            // Even 
            else
                count_even++; 
        
      
        // Condition check 
        if (count_odd % 2 == 1 && count_even % 2 == 1
            return false
      
        else
            return true
    
      
    // Drivers code 
    public static void main (String[] args) 
    
        int []a = { 1, 0, 1, 1, 0, 1 }; 
        int n = a.length; 
      
        if (flipsPossible(a, n)) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    
}
  
// This code is contributed by AnkitRai01

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
# Function to check if parity 
# can be made same or not 
def flipsPossible(a, n) : 
  
    count_odd = 0; count_even = 0
  
    # Iterate and count the parity 
    for i in range(n) :
  
        # Odd 
        if (a[i] & 1) :
            count_odd += 1
  
        # Even 
        else :
            count_even += 1
  
    # Condition check 
    if (count_odd % 2 and count_even % 2) :
        return False
    else :
        return True
  
# Driver Code 
if __name__ == "__main__"
  
    a = [ 1, 0, 1, 1, 0, 1 ]; 
      
    n = len(a); 
  
    if (flipsPossible(a, n)) :
        print("Yes"); 
    else :
        print("No"); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach 
using System;
  
class GFG 
{
      
    // Function to check if parity 
    // can be made same or not 
    static bool flipsPossible(int []a, int n) 
    
      
        int count_odd = 0, count_even = 0; 
      
        // Iterate and count the parity 
        for (int i = 0; i < n; i++) 
        
      
            // Odd 
            if ((a[i] & 1) == 1) 
                count_odd++; 
      
            // Even 
            else
                count_even++; 
        
      
        // Condition check 
        if (count_odd % 2 == 1 && count_even % 2 == 1) 
            return false
      
        else
            return true
    
      
    // Drivers code 
    public static void Main(String[] args) 
    
        int []a = { 1, 0, 1, 1, 0, 1 }; 
        int n = a.Length; 
      
        if (flipsPossible(a, n)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    
}
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

Yes



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, 29AjayKumar