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 Svoid 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 Codeint 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 approachclass GFG{// Function to count the total// number of gates required to// realize the boolean expression Sstatic 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 Codepublic 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 Sdef 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 Codeif __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 approachusing System;public class GFG{// Function to count the total// number of gates required to// realize the boolean expression Sstatic 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 Codepublic 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 Sfunction 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)
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