Minimum number of basic logic gates required to realize given Boolean expression

0 0 3 minutes read

Given a string S of length N representing a boolean expression, the task is to find the minimum number of AND, OR, and NOT gates required to realize the given expression.

Examples:

Input: S = “A+B.C”
Output: 2
Explanation: Realizing the expression requires 1 AND gate represented by ‘.’ and 1 OR gate represented by ‘+’.

Input: S = “(1 – A). B+C”
Output: 3
Explanation: Realizing the expression requires 1 AND gate represented by ‘.’ and 1 OR gate represented by ‘+’ and 1 NOT gate represented by ‘-‘.

Approach: Follow the steps below to solve the problem:

Iterate over the characters of the string.
Initialize, count of gates to 0.
If the current character is either ‘.’ or ‘+’, or ‘1’, then increment the count of gates by 1
Print the count of gates required.

Below is the implementation of the above approach:

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

Time Complexity: O(N)
Auxiliary Space: O(1)

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!