Número mínimo de bits necesarios para invertir de modo que Bitwise OR de A y B sea igual a C

Posted on by Rudeus Greyrat

Dados tres enteros positivos A , B y C , la tarea es contar el número mínimo de cambios de bits requeridos en A y B , de modo que el OR de A y B sea igual a C o no.

Ejemplos:

Entrada: A = 2, B = 2, C = 3
Salida: 1
Explicación:
La representación binaria de A es 010, B es 010 y C es 011. 
Voltee el tercer bit de A o B, de modo que A | B = C, es decir, 011 | 010 = 011.
Por lo tanto, el número total de lanzamientos necesarios es 1.

Entrada: A = 2, B = 6, C = 5
Salida: 3

Enfoque: siga los pasos a continuación para resolver el problema:

  • Inicialice una variable, digamos res , que almacene el número mínimo de bits volteados requeridos.
  • Iterar sobre cada bit de A , B y C y realizar los siguientes pasos:
    • Si el i -ésimo bit de C no está configurado, verifique lo siguiente:
      • Si se establece el i -ésimo bit de   A , incremente res con 1 , el i -ésimo bit de A debe invertirse.
      • Si se establece el i -ésimo bit de   B , incremente res con 1 ,   es necesario invertir el i -ésimo bit de B.
    • Si el i -ésimo bit de C está configurado, verifique lo siguiente:
      • Si el i -ésimo bit de A y B no está configurado, incremente res con 1 , cualquiera de los bits debe invertirse.
      • Si se establece el i -ésimo bit de A y B , no tenemos que cambiar ninguno de los bits.
  • Después de completar los pasos anteriores, imprima el valor de res como resultado.

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of
// bit flips required on A and B
// such that Bitwise OR of A and B is C
int minimumFlips(int A, int B, int C)
{
    // Stores the count of flipped bit
    int res = 0;
 
    // Iterate over the range [0, 32]
    for (int i = 0; i < 32; i++) {
 
        int x = 0, y = 0, z = 0;
 
        // Check if i-th bit of A is set
        if (A & (1 << i)) {
            x = 1;
        }
 
        // Check if i-th bit of B is
        // set or not
        if (B & (1 << i)) {
            y = 1;
        }
 
        // Check if i-th bit of C is
        // set or not
        if (C & (1 << i)) {
            z = 1;
        }
 
        // If i-th bit of C is unset
        if (z == 0) {
 
            // Check if i-th bit of
            // A is set or not
            if (x) {
                res++;
            }
 
            // Check if i-th bit of
            // B is set or not
            if (y) {
                res++;
            }
        }
 
        // Check if i-th bit of C
        // is set or not
        if (z == 1) {
 
            // Check if i-th bit of
            // both A and B is set
            if (x == 0 && y == 0) {
                res++;
            }
        }
    }
 
    // Return the count
    // of bits flipped
    return res;
}
 
// Driver Code
int main()
{
    int A = 2, B = 6, C = 5;
    cout << minimumFlips(A, B, C);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to count the number of
// bit flips required on A and B
// such that Bitwise OR of A and B is C
static int minimumFlips(int A, int B, int C)
{
     
    // Stores the count of flipped bit
    int res = 0;
 
    // Iterate over the range [0, 32]
    for(int i = 0; i < 32; i++)
    {
        int x = 0, y = 0, z = 0;
 
        // Check if i-th bit of A is set
        if ((A & (1 << i)) != 0)
        {
            x = 1;
        }
 
        // Check if i-th bit of B is
        // set or not
        if ((B & (1 << i)) != 0)
        {
            y = 1;
        }
 
        // Check if i-th bit of C is
        // set or not
        if ((C & (1 << i)) != 0)
        {
            z = 1;
        }
 
        // If i-th bit of C is unset
        if (z == 0)
        {
 
            // Check if i-th bit of
            // A is set or not
            if (x == 1)
            {
                res++;
            }
 
            // Check if i-th bit of
            // B is set or not
            if (y == 1)
            {
                res++;
            }
        }
 
        // Check if i-th bit of C
        // is set or not
        if (z == 1)
        {
 
            // Check if i-th bit of
            // both A and B is set
            if (x == 0 && y == 0)
            {
                res++;
            }
        }
    }
 
    // Return the count
    // of bits flipped
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int A = 2, B = 6, C = 5;
     
    System.out.println(minimumFlips(A, B, C));
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to count the number of
# bit flips required on A and B
# such that Bitwise OR of A and B is C
def minimumFlips(A, B, C):
     
    # Stores the count of flipped bit
    res = 0
 
    # Iterate over the range [0, 32]
    for i in range(32):
        x, y, z = 0, 0, 0
 
        # Check if i-th bit of A is set
        if (A & (1 << i)):
            x = 1
 
        # Check if i-th bit of B is
        # set or not
        if (B & (1 << i)):
            y = 1
 
        # Check if i-th bit of C is
        # set or not
        if (C & (1 << i)):
            z = 1
 
        # If i-th bit of C is unset
        if (z == 0):
             
            # Check if i-th bit of
            # A is set or not
            if (x):
                res += 1
 
            # Check if i-th bit of
            # B is set or not
            if (y):
                res += 1
 
        # Check if i-th bit of C
        # is set or not
        if (z == 1):
             
            # Check if i-th bit of
            # both A and B is set
            if (x == 0 and y == 0):
                res += 1
 
    # Return the count
    # of bits flipped
    return res
 
# Driver Code
if __name__ == '__main__':
     
    A, B, C = 2, 6, 5
     
    print(minimumFlips(A, B, C))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG {
 
    // Function to count the number of
    // bit flips required on A and B
    // such that Bitwise OR of A and B is C
    static int minimumFlips(int A, int B, int C)
    {
 
        // Stores the count of flipped bit
        int res = 0;
 
        // Iterate over the range [0, 32]
        for (int i = 0; i < 32; i++) {
            int x = 0, y = 0, z = 0;
 
            // Check if i-th bit of A is set
            if ((A & (1 << i)) != 0) {
                x = 1;
            }
 
            // Check if i-th bit of B is
            // set or not
            if ((B & (1 << i)) != 0) {
                y = 1;
            }
 
            // Check if i-th bit of C is
            // set or not
            if ((C & (1 << i)) != 0) {
                z = 1;
            }
 
            // If i-th bit of C is unset
            if (z == 0) {
 
                // Check if i-th bit of
                // A is set or not
                if (x == 1) {
                    res++;
                }
 
                // Check if i-th bit of
                // B is set or not
                if (y == 1) {
                    res++;
                }
            }
 
            // Check if i-th bit of C
            // is set or not
            if (z == 1) {
 
                // Check if i-th bit of
                // both A and B is set
                if (x == 0 && y == 0) {
                    res++;
                }
            }
        }
 
        // Return the count
        // of bits flipped
        return res;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int A = 2, B = 6, C = 5;
 
        Console.WriteLine(minimumFlips(A, B, C));
    }
}
 
// This code is contributed by ukasp.

Javascript

<script>
// javascript program for the above approach
    // Function to count the number of
    // bit flips required on A and B
    // such that Bitwise OR of A and B is C
    function minimumFlips(A , B , C) {
 
        // Stores the count of flipped bit
        var res = 0;
 
        // Iterate over the range [0, 32]
        for (i = 0; i < 32; i++) {
            var x = 0, y = 0, z = 0;
 
            // Check if i-th bit of A is set
            if ((A & (1 << i)) != 0) {
                x = 1;
            }
 
            // Check if i-th bit of B is
            // set or not
            if ((B & (1 << i)) != 0) {
                y = 1;
            }
 
            // Check if i-th bit of C is
            // set or not
            if ((C & (1 << i)) != 0) {
                z = 1;
            }
 
            // If i-th bit of C is unset
            if (z == 0) {
 
                // Check if i-th bit of
                // A is set or not
                if (x == 1) {
                    res++;
                }
 
                // Check if i-th bit of
                // B is set or not
                if (y == 1) {
                    res++;
                }
            }
 
            // Check if i-th bit of C
            // is set or not
            if (z == 1) {
 
                // Check if i-th bit of
                // both A and B is set
                if (x == 0 && y == 0) {
                    res++;
                }
            }
        }
 
        // Return the count
        // of bits flipped
        return res;
    }
 
    // Driver Code
        var A = 2, B = 6, C = 5;
        document.write(minimumFlips(A, B, C));
 
// This code is contributed by aashish1995
</script>
Producción: 

3

 

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Publicación traducida automáticamente

Artículo escrito por nirajgusain5 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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