Dada una array arr[][] de dimensiones M*N , la tarea es convertir cada elemento de la array en Bitwise XOR de dígitos presentes en el elemento.
Ejemplos:
Entrada: arr[][] = {{27, 173}, {5, 21}}
Salida:
5 5
5 3
Explicación:
Bitwise XOR de dígitos de arr[0][0] (= 27) es 5 (2^ 7).
El valor XOR bit a bit de los dígitos de arr[0][1] (= 173) es 5 (1 ^ 7 ^ 3).
El valor XOR bit a bit de los dígitos de arr[1][0] (= 5) es 5. El
valor XOR bit a bit de los dígitos de arr[1][1] (= 21) es 3(1 ^ 2).Entrada: arr[][] = {{11, 12, 33}, {64, 57, 61}, {74, 88, 39}} Salida: 0 3 0
2
2
7
3 0 10
Enfoque: Para resolver el problema, la idea del enfoque es atravesar la array dada y, para cada elemento de la array, imprimir el XOR bit a bit de sus dígitos .
A continuación se muestra la implementación del enfoque anterior:
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 X
int 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 digits
void 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 matrix
void 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 Code
int main()
{
int arr[][3] = { { 27, 173, 5 },
{ 21, 6, 624 },
{ 5, 321, 49 } };
convertXOR(arr);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG{
static int M = 3;
static int N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
static 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 digits
static 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 matrix
static 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 Code
public 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 approach M = 3 N = 3 # Function to calculate Bitwise # XOR of digits present in X def 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 digits def 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 matrix def 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 Code if __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 approach
using System;
using System.Collections.Generic;
class GFG{
static int M = 3;
static int N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
static 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 digits
static 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 matrix
static 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 Code
static 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 approach
let M = 3;
let N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
function 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 digits
function 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 matrix
function 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>
Producción:
5 5 5 3 6 0 5 0 13
Complejidad Temporal: O(M*N*log 10 K) donde K es el elemento máximo presente en la array .
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por ajaykr00kj y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA