# 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 ` `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

 ``

Output:

`3`

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

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