# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` ` `  `    ``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 `

Output:

```Yes
```

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Improved By : AnkitRai01, 29AjayKumar