Número mínimo de puertas lógicas básicas requeridas para realizar la expresión booleana dada

Dada una string S de longitud N que representa una expresión booleana , la tarea es encontrar el número mínimo de compuertas AND, OR y NOT necesarias para realizar la expresión dada.

Ejemplos:

Entrada: S = “A+BC”
Salida: 2
Explicación: Realizar la expresión requiere 1 compuerta AND representada por ‘.’ y 1 compuerta OR representada por ‘+’. 

Entrada: S = “(1 – A). B+C”
Salida: 3
Explicación: Realizar la expresión requiere 1 compuerta AND representada por ‘.’ y 1 puerta OR representada por ‘+’ y 1 puerta NOT representada por ‘-‘. 

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

1. Iterar sobre los caracteres de la string .
2. Inicializar, cuenta de puertas a 0 .
3. Si el carácter actual es ‘.’ o ‘+’ , o ‘1’ , luego incremente el conteo de puertas en 1
4. Imprime el conteo de puertas requeridas.

A continuación se muestra la implementación del enfoque anterior:

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the total
// number of gates required to
// realize the boolean expression S
void numberOfGates(string s)
{
    // Length of the string
    int N = s.size();
 
    // Stores the count
    // of total gates
    int ans = 0;
 
    // Traverse the string
    for (int i = 0; i < (int)s.size(); i++) {
 
        // AND, OR and NOT Gate
        if (s[i] == '.' || s[i] == '+'
            || s[i] == '1') {
            ans++;
        }
    }
 
    // Print the count
    // of gates required
    cout << ans;
}
 
// Driver Code
int main()
{
    // Input
    string S = "(1-A).B+C";
 
    // Function call to count the
    // total number of gates required
    numberOfGates(S);
}

// Java implementation of
// the above approach
class GFG{
 
// Function to count the total
// number of gates required to
// realize the boolean expression S
static void numberOfGates(String s)
{
     
    // Length of the string
    int N = s.length();
 
    // Stores the count
    // of total gates
    int ans = 0;
 
    // Traverse the string
    for(int i = 0; i < (int)s.length(); i++)
    {
         
        // AND, OR and NOT Gate
        if (s.charAt(i) == '.' ||
            s.charAt(i) == '+' ||
            s.charAt(i) == '1')
        {
            ans++;
        }
    }
 
    // Print the count
    // of gates required
    System.out.println(ans);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    String S = "(1-A).B+C";
 
    // Function call to count the
    // total number of gates required
    numberOfGates(S);
}
}
 
// This code is contributed by user_qa7r

# Python3 implementation of
# the above approach
 
# Function to count the total
# number of gates required to
# realize the boolean expression S
def numberOfGates(s):
 
    # Length of the string
    N = len(s)
 
    # Stores the count
    # of total gates
    ans = 0
 
    # Traverse the string
    for i in range(len(s)):
 
        # AND, OR and NOT Gate
        if (s[i] == '.' or s[i] == '+' or
            s[i] == '1'):
            ans += 1
 
    # Print the count
    # of gates required
    print(ans, end = "")
 
# Driver Code
if __name__ == "__main__":
 
    # Input
    S = "(1-A).B+C"
 
    # Function call to count the
    # total number of gates required
    numberOfGates(S)
 
# This code is contributed by AnkThon

// C# implementation of
// the above approach
using System;
public class GFG
{
 
// Function to count the total
// number of gates required to
// realize the boolean expression S
static void numberOfGates(string s)
{
     
    // Length of the string
    int N = s.Length;
 
    // Stores the count
    // of total gates
    int ans = 0;
 
    // Traverse the string
    for(int i = 0; i < s.Length; i++)
    {
         
        // AND, OR and NOT Gate
        if (s[i] == '.' ||
            s[i] == '+' ||
            s[i] == '1')
        {
            ans++;
        }
    }
 
    // Print the count
    // of gates required
    Console.WriteLine(ans);
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Input
    string S = "(1-A).B+C";
 
    // Function call to count the
    // total number of gates required
    numberOfGates(S);
}
}
 
// This code is contributed by AnkThon

<script>
// JavaScript program for the above approach
 
// Function to count the total
// number of gates required to
// realize the boolean expression S
function numberOfGates(s)
{
      
    // Length of the string
    let N = s.length;
  
    // Stores the count
    // of total gates
    let ans = 0;
  
    // Traverse the string
    for(let i = 0; i < s.length; i++)
    {
          
        // AND, OR and NOT Gate
        if (s[i] == '.' ||
            s[i] == '+' ||
            s[i] == '1')
        {
            ans++;
        }
    }
  
    // Print the count
    // of gates required
    document.write(ans);
}
 
// Driver Code
 
    // Input
    let S = "(1-A).B+C";
  
    // Function call to count the
    // total number of gates required
    numberOfGates(S);
     
</script>

3

Complejidad temporal: O(N)
Espacio auxiliar: O(1)