Dadas dos strings s1 y s2 . La tarea es tomar un carácter de la primera string y un carácter de la segunda string y verificar si los valores ASCII de ambos caracteres tienen el mismo número de bits establecidos. Imprime el número total de dichos pares.
Ejemplos:
Entrada: s1 = “xcd”, s2 = “swa”
Salida: 1 El
único par válido es (d, a) con valores ASCII como 100 y 97 respectivamente.
Ambos contienen 3 bits establecidos.
Entrada: s1 = «geeks», s2 = «forgeeks»
Salida: 17
Acercarse:
- Cree dos arrays arr1 y arr2 de tamaño 6 con todos los valores inicializados en 0 para almacenar la frecuencia del número de bits establecidos. Dado que el número máximo de bits establecidos en alfabetos en minúsculas es 6.
- Recorre la string s1 y encuentra el valor ASCII de cada carácter. Almacene la frecuencia del número de bits establecidos de cada valor ASCII en una array arr1. (Por ejemplo, si hay 3 caracteres con 4 bits establecidos, almacene 3 en arr[4])
- Realice una operación similar para la string s2 y almacene su valor en otra array arr2.
- Inicialice una variable de conteo con 0.
- Para el número total de pares, siga agregando (arr1[i] * arr2[i]) en la variable de conteo para todos los valores válidos de i.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of valid pairs
int totalPairs(string s1, string s2)
{
int count = 0;
int arr1[7], arr2[7];
// Initialise both arrays with 0
for (int i = 1; i <= 6; i++) {
arr1[i] = 0;
arr2[i] = 0;
}
// Store frequency of number of set bits for s1
for (int i = 0; i < s1.length(); i++) {
int set_bits = __builtin_popcount((int)s1[i]);
arr1[set_bits]++;
}
// Store frequency of number of set bits for s2
for (int i = 0; i < s2.length(); i++) {
int set_bits = __builtin_popcount((int)s2[i]);
arr2[set_bits]++;
}
// Calculate total pairs
for (int i = 1; i <= 6; i++)
count += (arr1[i] * arr2[i]);
// Return the count of valid pairs
return count;
}
// Driver code
int main()
{
string s1 = "geeks";
string s2 = "forgeeks";
cout << totalPairs(s1, s2);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the count of valid pairs
static int totalPairs(String s1, String s2)
{
int count = 0;
int[] arr1 = new int[7];
int[] arr2 = new int[7];
// Default Initialise both arrays 0
// Store frequency of number of set bits for s1
for (int i = 0; i < s1.length(); i++)
{
int set_bits = Integer.bitCount(s1.charAt(i));
arr1[set_bits]++;
}
// Store frequency of number of set bits for s2
for (int i = 0; i < s2.length(); i++)
{
int set_bits = Integer.bitCount(s2.charAt(i));
arr2[set_bits]++;
}
// Calculate total pairs
for (int i = 1; i <= 6; i++)
{
count += (arr1[i] * arr2[i]);
}
// Return the count of valid pairs
return count;
}
// Driver code
public static void main(String[] args)
{
String s1 = "geeks";
String s2 = "forgeeks";
System.out.println(totalPairs(s1, s2));
}
}
// This code has been contributed by 29AjayKumar
Python3
# Python3 implementation of the approach # Function to get no of set bits in binary # representation of positive integer n def countSetBits(n): count = 0 while (n): count += n & 1 n >>= 1 return count # Function to return the count # of valid pairs def totalPairs(s1, s2) : count = 0; arr1 = [0] * 7; arr2 = [0] * 7; # Store frequency of number # of set bits for s1 for i in range(len(s1)) : set_bits = countSetBits(ord(s1[i])) arr1[set_bits] += 1; # Store frequency of number of # set bits for s2 for i in range(len(s2)) : set_bits = countSetBits(ord(s2[i])); arr2[set_bits] += 1; # Calculate total pairs for i in range(1, 7) : count += (arr1[i] * arr2[i]); # Return the count of valid pairs return count; # Driver code if __name__ == "__main__" : s1 = "geeks"; s2 = "forgeeks"; print(totalPairs(s1, s2)); # This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
// Function to return the count of valid pairs
static int totalPairs(string s1, string s2)
{
int count = 0;
int[] arr1 = new int[7];
int[] arr2 = new int[7];
// Default Initialise both arrays 0
// Store frequency of number of set bits for s1
for (int i = 0; i < s1.Length; i++)
{
int set_bits = Convert.ToString((int)s1[i], 2).Count(c => c == '1');
arr1[set_bits]++;
}
// Store frequency of number of set bits for s2
for (int i = 0; i < s2.Length; i++)
{
int set_bits = Convert.ToString((int)s2[i], 2).Count(c => c == '1');
arr2[set_bits]++;
}
// Calculate total pairs
for (int i = 1; i <= 6; i++)
count += (arr1[i] * arr2[i]);
// Return the count of valid pairs
return count;
}
// Driver code
static void Main()
{
string s1 = "geeks";
string s2 = "forgeeks";
Console.WriteLine(totalPairs(s1, s2));
}
}
// This code is contributed by chandan_jnu
Javascript
<script>
// JavaScript implementation of the approach
// Function to get no of set bits in binary
// representation of positive integer n
function countSetBits(n) {
var count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
// Function to return the count
// of valid pairs
function totalPairs(s1, s2) {
var count = 0;
var arr1 = new Array(7).fill(0);
var arr2 = new Array(7).fill(0);
// Store frequency of number
// of set bits for s1
for (let i = 0; i < s1.length; i++) {
set_bits = countSetBits(s1[i].charCodeAt(0));
arr1[set_bits] += 1;
}
// Store frequency of number of
// set bits for s2
for (let i = 0; i < s2.length; i++) {
set_bits = countSetBits(s2[i].charCodeAt(0));
arr2[set_bits] += 1;
}
// Calculate total pairs
for (let i = 1; i < 7; i++) {
count += arr1[i] * arr2[i];
}
// Return the count of valid pairs
return count;
}
// Driver code
var s1 = "geeks";
var s2 = "forgeeks";
document.write(totalPairs(s1, s2));
</script>
Producción:
17
Publicación traducida automáticamente
Artículo escrito por Shashank_Sharma y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA