Evaluate a boolean expression represented as string
Last Updated :
15 Jul, 2022
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.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int evaluateBoolExpr(string s)
{
int n = s.length();
for ( int i = 0; i < n; i += 2) {
if (s[i + 1] == 'A' ) {
if (s[i + 2] == '0' || s[i] == '0' )
s[i + 2] = '0' ;
else
s[i + 2] = '1' ;
}
else if (s[i + 1] == 'B' ) {
if (s[i + 2] == '1' || s[i] == '1' )
s[i + 2] = '1' ;
else
s[i + 2] = '0' ;
}
else {
if (s[i + 2] == s[i])
s[i + 2] = '0' ;
else
s[i + 2] = '1' ;
}
}
return s[n - 1] - '0' ;
}
int main()
{
string s = "1C1B1B0A0" ;
cout << evaluateBoolExpr(s);
return 0;
}
|
Java
public class Evaluate_BoolExp {
static int evaluateBoolExpr(StringBuffer s)
{
int n = s.length();
for ( int i = 0 ; i < n; i += 2 ) {
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' );
}
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' );
}
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' ;
}
public static void main(String[] args)
{
String s = "1C1B1B0A0" ;
StringBuffer sb = new StringBuffer(s);
System.out.println(evaluateBoolExpr(sb));
}
}
|
Python3
import math as mt
def evaluateBoolExpr(s):
n = len (s)
for i in range ( 0 , n - 2 , 2 ):
if (s[i + 1 ] = = "A" ):
if (s[i + 2 ] = = "0" or s[i] = = "0" ):
s[i + 2 ] = "0"
else :
s[i + 2 ] = "1"
else if (s[i + 1 ] = = "B" ):
if (s[i + 2 ] = = "1" or s[i] = = "1" ):
s[i + 2 ] = "1"
else :
s[i + 2 ] = "0"
else :
if (s[i + 2 ] = = s[i]):
s[i + 2 ] = "0"
else :
s[i + 2 ] = "1"
return ord (s[n - 1 ]) - ord ( "0" )
s = "1C1B1B0A0"
string = [s[i] for i in range ( len (s))]
print (evaluateBoolExpr(string))
|
C#
using System;
using System.Text;
class GFG
{
public static int evaluateBoolExpr(StringBuilder s)
{
int n = s.Length;
for ( int i = 0; i < n; i += 2)
{
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' ;
}
}
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' ;
}
}
else
{
if (s[i + 2] == s[i])
{
s[i + 2] = '0' ;
}
else
{
s[i + 2] = '1' ;
}
}
}
}
return s[n - 1] - '0' ;
}
public static void Main( string [] args)
{
string s = "1C1B1B0A0" ;
StringBuilder sb = new StringBuilder(s);
Console.WriteLine(evaluateBoolExpr(sb));
}
}
|
PHP
<?php
function evaluateBoolExpr( $s )
{
$n = strlen ( $s );
for ( $i = 0; $i < $n ; $i += 2)
{
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' ;
}
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' ;
}
else
{
if (( $i + 2) < $n && $s [ $i + 2] == $s [ $i ])
$s [ $i + 2] = '0' ;
else
$s [ $i + 2] = '1' ;
}
}
return $s [ $n - 1] - '0' ;
}
$s = "1C1B1B0A0" ;
echo evaluateBoolExpr( $s );
|
Javascript
<script>
function evaluateBoolExpr(s)
{
let n = s.length;
for (let i = 0; i < n; i += 2)
{
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' ;
}
}
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' ;
}
}
else
{
if (s[i + 2] == s[i])
{
s[i + 2] = '0' ;
}
else
{
s[i + 2] = '1' ;
}
}
}
}
return (s[n - 1].charCodeAt() -
'0' .charCodeAt());
}
let s = "1C1B1B0A0" ;
let sb = s.split( '' );
document.write(evaluateBoolExpr(sb) + "</br>" );
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
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