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

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:

1. Iterate over the characters of the string.
2. Initialize, count of gates to 0.
3. If the current character is either ‘.’ or ‘+’, or ‘1’, then increment the count of gates by 1
4. 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:

3

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