Given a matrix mat[][] of size N * M, the task is to check if it is possible to rearrange the row elements of the matrix such that Bitwise XOR of the first column element is non-zero. If it is possible then print “Yes” else print “No”.
Examples:
Input: mat[][] = {{1, 1, 2}, {2, 2, 2}, {3, 3, 3}}
Output: Yes
Explanation:
After rearranging the first row as 2, 1, 1.
Bitwise XOR of the first column will be 3 i.e., (2 ^ 2 ^ 3).Input: mat[][] = {{1, 1, 1}, {2, 2, 2}, {3, 3, 3}}
Output: No
Explanation:
As all the rearrangements give same first element so the only combination is (1 ^ 2 ^ 3) which equals zero.
Therefore, it is not possible to obtain non-zero Bitwise XOR of the first column.
Approach: Follow the steps below to solve the problem:
- Find the Bitwise XOR of the elements of the first column of the matrix and store it in a variable res.
- If res is non-zero then print “Yes”.
- Else, iterate over all the rows and find the element in a row which is not equal to the element at the first index of this row.
- If no such element is present in any row in the above step then print “No” else then print “Yes”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if there is any// row where number of unique elements// are greater than 1string checkRearrangements( vector<vector<int> > mat, int N, int M){ // Iterate over the matrix for (int i = 0; i < N; i++) { for (int j = 1; j < M; j++) { if (mat[i][0] != mat[i][j]) { return "Yes"; } } } return "No";}// Function to check if it is possible// to rearrange mat[][] such that XOR// of its first column is non-zerostring nonZeroXor(vector<vector<int> > mat, int N, int M){ int res = 0; // Find bitwise XOR of the first // column of mat[][] for (int i = 0; i < N; i++) { res = res ^ mat[i][0]; } // If bitwise XOR of the first // column of mat[][] is non-zero if (res != 0) return "Yes"; // Otherwise check rearrangements else return checkRearrangements(mat, N, M);}// Driver Codeint main(){ // Given Matrix mat[][] vector<vector<int> > mat = { { 1, 1, 2 }, { 2, 2, 2 }, { 3, 3, 3 } }; int N = mat.size(); int M = mat[0].size(); // Function Call cout << nonZeroXor(mat, N, M); return 0;} |
Java
// Java program for the // above approachimport java.util.*;class GFG{// Function to check if there is any// row where number of unique elements// are greater than 1static String checkRearrangements(int[][] mat, int N, int M){ // Iterate over the matrix for (int i = 0; i < N; i++) { for (int j = 1; j < M; j++) { if (mat[i][0] != mat[i][j]) { return "Yes"; } } } return "No";}// Function to check if it is possible// to rearrange mat[][] such that XOR// of its first column is non-zerostatic String nonZeroXor(int[][] mat, int N, int M){ int res = 0; // Find bitwise XOR of the // first column of mat[][] for (int i = 0; i < N; i++) { res = res ^ mat[i][0]; } // If bitwise XOR of the first // column of mat[][] is non-zero if (res != 0) return "Yes"; // Otherwise check // rearrangements else return checkRearrangements(mat, N, M);}// Driver Codepublic static void main(String[] args){ // Given Matrix mat[][] int[][] mat = {{1, 1, 2}, {2, 2, 2}, {3, 3, 3}}; int N = mat.length; int M = mat[0].length; // Function Call System.out.print(nonZeroXor(mat, N, M));}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program for the above approach# Function to check if there is any# row where number of unique elements# are greater than 1def checkRearrangements(mat, N, M): # Iterate over the matrix for i in range(N): for j in range(1, M): if (mat[i][0] != mat[i][j]): return "Yes" return "No"# Function to check if it is possible# to rearrange mat[][] such that XOR# of its first column is non-zerodef nonZeroXor(mat, N, M): res = 0 # Find bitwise XOR of the first # column of mat[][] for i in range(N): res = res ^ mat[i][0] # If bitwise XOR of the first # column of mat[][] is non-zero if (res != 0): return "Yes" # Otherwise check rearrangements else: return checkRearrangements(mat, N, M)# Driver Codeif __name__ == "__main__": # Given Matrix mat[][] mat = [ [ 1, 1, 2 ], [ 2, 2, 2 ], [ 3, 3, 3 ] ] N = len(mat) M = len(mat[0]) # Function Call print(nonZeroXor(mat, N, M))# This code is contributed by chitranayal |
C#
// C# program for the // above approachusing System;class GFG{// Function to check if there is any// row where number of unique elements// are greater than 1static String checkRearrangements(int[,] mat, int N, int M){ // Iterate over the matrix for(int i = 0; i < N; i++) { for(int j = 1; j < M; j++) { if (mat[i, 0] != mat[i, j]) { return "Yes"; } } } return "No";}// Function to check if it is possible// to rearrange [,]mat such that XOR// of its first column is non-zerostatic String nonZeroXor(int[,] mat, int N, int M){ int res = 0; // Find bitwise XOR of the // first column of [,]mat for(int i = 0; i < N; i++) { res = res ^ mat[i, 0]; } // If bitwise XOR of the first // column of [,]mat is non-zero if (res != 0) return "Yes"; // Otherwise check // rearrangements else return checkRearrangements(mat, N, M);}// Driver Codepublic static void Main(String[] args){ // Given Matrix [,]mat int[,] mat = { { 1, 1, 2 }, { 2, 2, 2 }, { 3, 3, 3 } }; int N = mat.GetLength(0); int M = mat.GetLength(1); // Function Call Console.Write(nonZeroXor(mat, N, M));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// JavaScript program to implement// the above approach// Function to check if there is any// row where number of unique elements// are greater than 1function checkRearrangements(mat, N, M){ // Iterate over the matrix for (let i = 0; i < N; i++) { for (let j = 1; j < M; j++) { if (mat[i][0] != mat[i][j]) { return "Yes"; } } } return "No";} // Function to check if it is possible// to rearrange mat[][] such that XOR// of its first column is non-zerofunction nonZeroXor(mat, N, M){ let res = 0; // Find bitwise XOR of the // first column of mat[][] for (let i = 0; i < N; i++) { res = res ^ mat[i][0]; } // If bitwise XOR of the first // column of mat[][] is non-zero if (res != 0) return "Yes"; // Otherwise check // rearrangements else return checkRearrangements(mat, N, M);}// Driver Code // Given Matrix mat[][] let mat = [[1, 1, 2], [2, 2, 2], [3, 3, 3]]; let N = mat.length; let M = mat[0].length; // Function Call document.write(nonZeroXor(mat, N, M)); </script> |
Output:
Yes
Time Complexity: O(N * M)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
