Check if all the elements can be made of same parity by inverting adjacent elements Last Updated : 22 Nov, 2021 Comments Improve Suggest changes Like Article Like Report 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> OutputYes Time Complexity: O(N) Auxiliary Space: O(1) Comment More infoAdvertise with us Next Article Check if all the elements can be made of same parity by inverting adjacent elements S Striver Follow Improve Article Tags : Analysis of Algorithms Matrix DSA Arrays binary-representation +1 More Practice Tags : ArraysMatrix Similar Reads Check if all elements of a Circular Array can be made equal by increments of adjacent pairs Given a circular array arr[] of size N, the task is to check if it is possible to make all array elements of the circular array equal by increasing pairs of adjacent elements by 1. 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