Modify a matrix by converting each element to XOR of its digits
Given a matrix arr[][] of dimensions M*N, the task is to convert every matrix element to Bitwise XOR of digits present in the element.
Examples:
Input: arr[][] = {{27, 173}, {5, 21}}
Output:
5 5
5 3
Explanation:
Bitwise XOR of digits of arr[0][0] (= 27) is 5 (2^7).
Bitwise XOR value of digits of arr[0][1] (= 173) is 5 (1 ^ 7 ^ 3).
Bitwise XOR value of digits of arr[1][0] (= 5) is 5.
Bitwise XOR value of digits of arr[1][1] (= 21) is 3(1 ^ 2).Input: arr[][] = {{11, 12, 33}, {64, 57, 61}, {74, 88, 39}}
Output:
0 3 0
2 2 7
3 0 10
Approach: To solve the problem, the approach idea is to traverse the given matrix and for each matrix element, print the Bitwise XOR of its digits.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int M = 3;const int N = 3;// Function to calculate Bitwise// XOR of digits present in Xint findXOR(int X){ // Stores the Bitwise XOR int ans = 0; // While X is true while (X) { // Update Bitwise // XOR of its digits ans ^= (X % 10); X /= 10; } // Return the result return ans;}// Function to print matrix after// converting each matrix element// to XOR of its digitsvoid printXORmatrix(int arr[M][N]){ // Traverse each row of arr[][] for (int i = 0; i < M; i++) { // Traverse each column of arr[][] for (int j = 0; j < N; j++) { cout << arr[i][j] << " "; } cout << "\n"; }}// Function to convert the given// matrix to required XOR matrixvoid convertXOR(int arr[M][N]){ // Traverse each row of arr[][] for (int i = 0; i < M; i++) { // Traverse each column of arr[][] for (int j = 0; j < N; j++) { // Store the current // matrix element int X = arr[i][j]; // Find the xor of // digits present in X int temp = findXOR(X); // Stores the XOR value arr[i][j] = temp; } } // Print resultant matrix printXORmatrix(arr);}// Driver Codeint main(){ int arr[][3] = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; convertXOR(arr); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{static int M = 3;static int N = 3;// Function to calculate Bitwise// XOR of digits present in Xstatic int findXOR(int X){ // Stores the Bitwise XOR int ans = 0; // While X is true while (X != 0) { // Update Bitwise // XOR of its digits ans ^= (X % 10); X /= 10; } // Return the result return ans;}// Function to print matrix after// converting each matrix element// to XOR of its digitsstatic void printXORmatrix(int arr[][]){ // Traverse each row of arr[][] for(int i = 0; i < M; i++) { // Traverse each column of arr[][] for(int j = 0; j < N; j++) { System.out.print(arr[i][j] + " "); } System.out.println(); }}// Function to convert the given// matrix to required XOR matrixstatic void convertXOR(int arr[][]){ // Traverse each row of arr[][] for(int i = 0; i < M; i++) { // Traverse each column of arr[][] for(int j = 0; j < N; j++) { // Store the current // matrix element int X = arr[i][j]; // Find the xor of // digits present in X int temp = findXOR(X); // Stores the XOR value arr[i][j] = temp; } } // Print resultant matrix printXORmatrix(arr);}// Driver Codepublic static void main (String[] args){ int arr[][] = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; convertXOR(arr);}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approachM = 3N = 3# Function to calculate Bitwise# XOR of digits present in Xdef findXOR(X): # Stores the Bitwise XOR ans = 0 # While X is true while (X): # Update Bitwise # XOR of its digits ans ^= (X % 10) X //= 10 # Return the result return ans# Function to print matrix after# converting each matrix element# to XOR of its digitsdef printXORmatrix(arr): # Traverse each row of arr[][] for i in range(3): # Traverse each column of arr[][] for j in range(3): print(arr[i][j], end = " ") print()# Function to convert the given# matrix to required XOR matrixdef convertXOR(arr): # Traverse each row of arr[][] for i in range(3): # Traverse each column of arr[][] for j in range(3): # Store the current # matrix element X = arr[i][j] # Find the xor of # digits present in X temp = findXOR(X) # Stores the XOR value arr[i][j] = temp # Print resultant matrix printXORmatrix(arr)# Driver Codeif __name__ == '__main__': arr=[[27, 173, 5], [ 21, 6, 624 ], [ 5, 321, 49 ]] convertXOR(arr)# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{static int M = 3;static int N = 3;// Function to calculate Bitwise// XOR of digits present in Xstatic int findXOR(int X){ // Stores the Bitwise XOR int ans = 0; // While X is true while (X != 0) { // Update Bitwise // XOR of its digits ans ^= (X % 10); X /= 10; } // Return the result return ans;}// Function to print matrix after// converting each matrix element// to XOR of its digitsstatic void printXORmatrix(int[,] arr){ // Traverse each row of arr[][] for(int i = 0; i < M; i++) { // Traverse each column of arr[][] for(int j = 0; j < N; j++) { Console.Write(arr[i, j] + " "); } Console.WriteLine(); }}// Function to convert the given// matrix to required XOR matrixstatic void convertXOR(int[,] arr){ // Traverse each row of arr[][] for(int i = 0; i < M; i++) { // Traverse each column of arr[][] for(int j = 0; j < N; j++) { // Store the current // matrix element int X = arr[i, j]; // Find the xor of // digits present in X int temp = findXOR(X); // Stores the XOR value arr[i, j] = temp; } } // Print resultant matrix printXORmatrix(arr);}// Driver Codestatic public void Main(){ int[,] arr = { { 27, 173, 5 }, { 21, 6, 624 }, { 5, 321, 49 } }; convertXOR(arr);}}// This code is contributed by splevel62 |
Javascript
<script>// JavaScript program to implement// the above approachlet M = 3;let N = 3; // Function to calculate Bitwise// XOR of digits present in Xfunction findXOR(X){ // Stores the Bitwise XOR let ans = 0; // While X is true while (X != 0) { // Update Bitwise // XOR of its digits ans ^= (X % 10); X /= 10; } // Return the result return ans;} // Function to print matrix after// converting each matrix element// to XOR of its digitsfunction printXORmatrix(arr){ // Traverse each row of arr[][] for(let i = 0; i < M; i++) { // Traverse each column of arr[][] for(let j = 0; j < N; j++) { document.write(arr[i][j] + " "); } document.write("<br/>"); }} // Function to convert the given// matrix to required XOR matrixfunction convertXOR(arr){ // Traverse each row of arr[][] for(let i = 0; i < M; i++) { // Traverse each column of arr[][] for(let j = 0; j < N; j++) { // Store the current // matrix element let X = arr[i][j]; // Find the xor of // digits present in X let temp = findXOR(X); // Stores the XOR value arr[i][j] = temp; } } // Print resultant matrix prletXORmatrix(arr);}// Driver code let arr = [[ 27, 173, 5 ], [ 21, 6, 624 ], [ 5, 321, 49 ]]; convertXOR(arr);;// This code is contributed by susmitakundugoaldanga.</script> |
Output:
5 5 5 3 6 0 5 0 13
Time Complexity: O(M*N*log10K) where K is the maximum element present in the matrix.
Auxiliary Space: O(1)
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