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Evaluate a boolean expression represented as string
  • Difficulty Level : Basic
  • Last Updated : 10 Jan, 2019
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Given a string consisting of only 0, 1, A, B, C where
A = AND
B = OR
C = XOR
Calculate the value of the string assuming no order of precedence and evaluation is done from left to right.

Constraints – The length of string will be odd. It will always be a valid string.
Example, 1AA0 will not be given as an input.

Examples:

Input : 1A0B1
Output : 1
1 AND 0 OR 1 = 1

Input : 1C1B1B0A0
Output : 0

Source : Microsoft online round for internship 2017



The idea is to traverse all operands by jumping a character after every iteration. For current operand str[i], check values of str[i+1] and str[i+2], accordingly decide the value of current subexpression.

C++




// C++ program to evaluate value of an expression.
#include <bits/stdc++.h>
  
using namespace std;
  
int evaluateBoolExpr(string s)
{
    int n = s.length();
  
    // Traverse all operands by jumping
    // a character after every iteration.
    for (int i = 0; i < n; i += 2) {
  
        // If operator next to current operand
        // is AND.
        if (s[i + 1] == 'A') {
            if (s[i + 2] == '0'|| s[i] == '0')
                s[i + 2] = '0';
            else
                s[i + 2] = '1';
        }
  
        // If operator next to current operand
        // is OR.
        else if (s[i + 1] == 'B') {
            if (s[i + 2] == '1'|| s[i] == '1')
                s[i + 2] = '1';
            else
                s[i + 2] = '0';
        }
  
        // If operator next to current operand
        // is XOR (Assuming a valid input)
        else {
            if (s[i + 2] == s[i])
                s[i + 2] = '0';
            else
                s[i + 2] = '1';
        }
    }
    return s[n - 1] -'0';
}
  
// Driver code
int main()
{
    string s = "1C1B1B0A0";
    cout << evaluateBoolExpr(s);
    return 0;
}

Java




// Java program to evaluate value of an expression.
public class Evaluate_BoolExp {
      
    // Evaluates boolean expression
    // and returns the result
    static int evaluateBoolExpr(StringBuffer s)
    {
        int n = s.length();
       
        // Traverse all operands by jumping
        // a character after every iteration.
        for (int i = 0; i < n; i += 2) {
       
            // If operator next to current operand
            // is AND.
            if( i + 1 < n && i + 2 < n)
            {
                if (s.charAt(i + 1) == 'A') {
                    if (s.charAt(i + 2) == '0' || 
                            s.charAt(i) == 0)
                        s.setCharAt(i + 2, '0');
                    else
                        s.setCharAt(i + 2, '1');
                }
           
                // If operator next to current operand
                // is OR.
                else if ((i + 1) < n && 
                           s.charAt(i + 1 ) == 'B') {
                    if (s.charAt(i + 2) == '1' ||
                          s.charAt(i) == '1')
                        s.setCharAt(i + 2, '1');
                    else
                        s.setCharAt(i + 2, '0');
                }
                  
                // If operator next to current operand
                // is XOR (Assuming a valid input)
                else {
                    if (s.charAt(i + 2) == s.charAt(i))
                        s.setCharAt(i + 2, '0');
                    else
                        s.setCharAt(i + 2 ,'1');
                }
            }
        }
        return s.charAt(n - 1) - '0';
    }
       
    // Driver code
    public static void main(String[] args) 
    {
        String s = "1C1B1B0A0";
        StringBuffer sb = new StringBuffer(s);
        System.out.println(evaluateBoolExpr(sb));
    }
}
// This code is contributed by Sumit Ghosh

Python3




# Python3 program to evaluate value 
# of an expression.
import math as mt
  
def evaluateBoolExpr(s):
  
    n = len(s)
      
    # Traverse all operands by jumping
    # a character after every iteration.
    for i in range(0, n - 2, 2):
  
        # If operator next to current 
        # operand is AND.'''
        if (s[i + 1] == "A"):
  
            if (s[i + 2] == "0" or s[i] == "0"):
                s[i + 2] = "0"
            else:
                s[i + 2] = "1"
  
        # If operator next to current 
        # operand is OR.
        elif (s[i + 1] == "B"):
            if (s[i + 2] == "1" or s[i] == "1"):
                s[i + 2] = "1"
            else:
                s[i + 2] = "0"
  
        # If operator next to current operand
        # is XOR (Assuming a valid input)
        else:
            if (s[i + 2] == s[i]):
                s[i + 2] = "0"
            else:
                s[i + 2] = "1"
  
    return ord(s[n - 1]) - ord("0")
  
# Driver code
s = "1C1B1B0A0"
string=[s[i] for i in range(len(s))]
print(evaluateBoolExpr(string))
  
# This code is contributed 
# by mohit kumar 29

C#




// C# program to evaluate value 
// of an expression. 
using System;
using System.Text;
  
class GFG
{
  
// Evaluates boolean expression 
// and returns the result 
public static int evaluateBoolExpr(StringBuilder s)
{
    int n = s.Length;
  
    // Traverse all operands by jumping 
    // a character after every iteration. 
    for (int i = 0; i < n; i += 2)
    {
  
        // If operator next to current 
        // operand is AND. 
        if (i + 1 < n && i + 2 < n)
        {
            if (s[i + 1] == 'A')
            {
                if (s[i + 2] == '0' || s[i] == 0)
                {
                    s[i + 2] = '0';
                }
                else
                {
                    s[i + 2] = '1';
                }
            }
  
            // If operator next to current 
            // operand is OR. 
            else if ((i + 1) < n && s[i + 1] == 'B')
            {
                if (s[i + 2] == '1' || s[i] == '1')
                {
                    s[i + 2] = '1';
                }
                else
                {
                    s[i + 2] = '0';
                }
            }
  
            // If operator next to current operand 
            // is XOR (Assuming a valid input) 
            else
            {
                if (s[i + 2] == s[i])
                {
                    s[i + 2] = '0';
                }
                else
                {
                    s[i + 2] = '1';
                }
            }
        }
    }
    return s[n - 1] - '0';
}
  
// Driver code 
public static void Main(string[] args)
{
    string s = "1C1B1B0A0";
    StringBuilder sb = new StringBuilder(s);
    Console.WriteLine(evaluateBoolExpr(sb));
}
}
  
// This code is contributed by Shrikant13

PHP




<?php
// PHP program to evaluate value 
// of an expression.
  
function evaluateBoolExpr($s)
{
    $n = strlen($s);
  
    // Traverse all operands by jumping
    // a character after every iteration.
    for ($i = 0; $i < $n; $i += 2)
    {
  
        // If operator next to current operand
        // is AND.
        if (($i + 1) < $n  && $s[$i + 1] == 'A')
        {
            if ($s[$i + 2] == '0'|| $s[$i] == '0')
                $s[$i + 2] = '0';
            else
                $s[$i + 2] = '1';
        }
  
        // If operator next to current operand
        // is OR.
        else if (($i + 1) < $n  && $s[$i + 1] == 'B')
        {
            if ($s[$i + 2] == '1'|| $s[$i] == '1')
                $s[$i + 2] = '1';
            else
                $s[$i + 2] = '0';
        }
  
        // If operator next to current operand
        // is XOR (Assuming a valid input)
        else 
        {
            if (($i + 2) < $n  && $s[$i + 2] == $s[$i])
                $s[$i + 2] = '0';
            else
                $s[$i + 2] = '1';
        }
    }
    return $s[$n - 1] -'0';
}
  
// Driver code
$s = "1C1B1B0A0";
echo evaluateBoolExpr($s);
  
// This code is contributed
// by Akanksha Rai


Output:
0

This article is contributed by Ayushi Jain. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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