Dadas dos strings binarias de longitudes desiguales A y B , la tarea es imprimir la string binaria que es el XOR de A y B.
Ejemplos:
Entrada: A = “11001”, B = “111111”
Salida: 100110
Entrada: A = “11111”, B = “0”
Salida: 11111
Enfoque: La idea es primero hacer que ambas strings tengan la misma longitud y luego realizar el XOR de cada carácter uno por uno y almacenarlo en la string resultante.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to insert n 0s in the
// beginning of the given string
void addZeros(string& str, int n)
{
for (int i = 0; i < n; i++) {
str = "0" + str;
}
}
// Function to return the XOR
// of the given strings
string getXOR(string a, string b)
{
// Lengths of the given strings
int aLen = a.length();
int bLen = b.length();
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen) {
addZeros(b, aLen - bLen);
}
else if (bLen > aLen) {
addZeros(a, bLen - aLen);
}
// Updated length
int len = max(aLen, bLen);
// To store the resultant XOR
string res = "";
for (int i = 0; i < len; i++) {
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
// Driver code
int main()
{
string a = "11001", b = "111111";
cout << getXOR(a, b);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to insert n 0s in the
// beginning of the given string
static String addZeros(String str, int n)
{
for (int i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
static String getXOR(String a, String b)
{
// Lengths of the given strings
int aLen = a.length();
int bLen = b.length();
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
int len = Math.max(aLen, bLen);
// To store the resultant XOR
String res = "";
for (int i = 0; i < len; i++)
{
if (a.charAt(i) == b.charAt(i))
res += "0";
else
res += "1";
}
return res;
}
// Driver code
public static void main (String[] args)
{
String a = "11001", b = "111111";
System.out.println(getXOR(a, b));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach # Function to insert n 0s in the # beginning of the given string def addZeros(strr, n): for i in range(n): strr = "0" + strr return strr # Function to return the XOR # of the given strings def getXOR(a, b): # Lengths of the given strings aLen = len(a) bLen = len(b) # Make both the strings of equal lengths # by inserting 0s in the beginning if (aLen > bLen): b = addZeros(b, aLen - bLen) elif (bLen > aLen): a = addZeros(a, bLen - aLen) # Updated length lenn = max(aLen, bLen); # To store the resultant XOR res = "" for i in range(lenn): if (a[i] == b[i]): res += "0" else: res += "1" return res # Driver code a = "11001" b = "111111" print(getXOR(a, b)) # This code is contributed by Mohit Kumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to insert n 0s in the
// beginning of the given string
static String addZeros(String str, int n)
{
for (int i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
static String getXOR(String a, String b)
{
// Lengths of the given strings
int aLen = a.Length;
int bLen = b.Length;
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
int len = Math.Max(aLen, bLen);
// To store the resultant XOR
String res = "";
for (int i = 0; i < len; i++)
{
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
// Driver code
public static void Main(String[] args)
{
String a = "11001", b = "111111";
Console.WriteLine(getXOR(a, b));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript implementation of the approach
// Function to insert n 0s in the
// beginning of the given string
function addZeros(str, n)
{
for (let i = 0; i < n; i++)
{
str = "0" + str;
}
return str;
}
// Function to return the XOR
// of the given strings
function getXOR(a, b)
{
// Lengths of the given strings
let aLen = a.length;
let bLen = b.length;
// Make both the strings of equal lengths
// by inserting 0s in the beginning
if (aLen > bLen)
{
a = addZeros(b, aLen - bLen);
}
else if (bLen > aLen)
{
a = addZeros(a, bLen - aLen);
}
// Updated length
let len = Math.max(aLen, bLen);
// To store the resultant XOR
let res = "";
for (let i = 0; i < len; i++)
{
if (a[i] == b[i])
res += "0";
else
res += "1";
}
return res;
}
let a = "11001", b = "111111";
document.write(getXOR(a, b));
// This code is contributed by divyeshrabadiya07.
</script>
Producción:
100110
Complejidad de tiempo: O(len), len=Max(longitud a,longitud b)
Espacio Auxiliar: O(len)