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

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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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>
 
// JavaScript implementation of the approach 
 
     
// Function to check if parity 
// can be made same or not 
function flipsPossible(a, n){
 
 
    let count_odd = 0;
    let count_even = 0;
 
    // Iterate and count the parity 
    for (let 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 
 
let a = [1, 0, 1, 1, 0, 1];
let n = a.length; 
 
if (flipsPossible(a, n)) 
    document.write("Yes");
else
    document.write("No");
     
</script>

Output

Yes

Time Complexity: O(N)

Auxiliary Space: O(1)

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