Solve the Logical Expression given by string - GeeksforGeeks

27 May, 2021
Given string str representing a logical expression which consists of the operators | (OR), & (AND),! (NOT) , 0, 1 and, only (i.e. no space between characters). The task is to print the result of the logical expression.Examples:  
Input: str = “[[0,&,1],|,[!,1]]” Output: 0 [[0,&,1],|,[!,1]] [[0,&,1],|,0] [[0,&,1],|,0] [0,|,0] [0,|,0] [0] 0Input: str = “[!,[[0,&,[!,1]],|,[!,[[!,0],&,1]]]]” Output: 1  
 
Approach:  
Start traversing the string from the end.If [ found go to Step-3 otherwise push the characters into the stack.Pop characters from the stack until the stack top become”]”. > Insert each popped character into the vector.If the stack top becomes ] after 5 pop operations then the vector will be x, |, y or x, &, y.If the stack top becomes ] after 3 pop operations then the vector will be !, x.Pop ] from the stack top.Perform the respective operations on the vector elements then push the result back into the stack.If the string is fully traversed then return the value at the stack top otherwise go to step 2.
Start traversing the string from the end.
If [ found go to Step-3 otherwise push the characters into the stack.
Pop characters from the stack until the stack top become”]”. > Insert each popped character into the vector.If the stack top becomes ] after 5 pop operations then the vector will be x, |, y or x, &, y.If the stack top becomes ] after 3 pop operations then the vector will be !, x.Pop ] from the stack top.
Pop characters from the stack until the stack top become”]”. > Insert each popped character into the vector.
If the stack top becomes ] after 5 pop operations then the vector will be x, |, y or x, &, y.
If the stack top becomes ] after 3 pop operations then the vector will be !, x.
Pop ] from the stack top.
Perform the respective operations on the vector elements then push the result back into the stack.
If the string is fully traversed then return the value at the stack top otherwise go to step 2.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// C++ program to solve the logical expression.#include <bits/stdc++.h>using namespace std; // Function to evaluate the logical expressionchar logicalExpressionEvaluation(string str){    stack<char> arr;     // traversing string from the end.    for (int i = str.length() - 1; i >= 0; i--)    {        if (str[i] == '[')        {            vector<char> s;            while (arr.top() != ']')            {                s.push_back(arr.top());                arr.pop();            }            arr.pop();             // for NOT operation            if (s.size() == 3)            {                s[2] == '1' ? arr.push('0') : arr.push('1');            }            // for AND and OR operation            else if (s.size() == 5)            {                int a = s[0] - 48, b = s[4] - 48, c;                s[2] == '&' ? c = a && b : c = a || b;                arr.push((char)c + 48);            }        }        else        {            arr.push(str[i]);        }    }    return arr.top();} // Driver codeint main(){    string str = "[[0,&,1],|,[!,1]]";     cout << logicalExpressionEvaluation(str) << endl;     return 0;}
// Java program to solve the logical expression.import java.util.*; class GFG{     // Function to evaluate the logical expressionstatic char logicalExpressionEvaluation(String str){    Stack<Character> arr = new Stack<Character>();     // traversing string from the end.    for (int i = str.length() - 1; i >= 0; i--)    {        if (str.charAt(i) == '[')        {            Vector<Character> s = new Stack<Character>();            while (arr.peek() != ']')            {                s.add(arr.peek());                arr.pop();            }            arr.pop();             // for NOT operation            if (s.size() == 3)            {                arr.push(s.get(2) == '1' ? '0' : '1');            }                         // for AND and OR operation            else if (s.size() == 5)            {                int a = s.get(0) - 48,                    b = s.get(4) - 48, c;                if(s.get(2) == '&' )                {                    c = a & b;                }                else                {                    c = a | b;                }                arr.push((char)(c + 48));            }        }        else        {            arr.push(str.charAt(i));        }    }    return arr.peek();} // Driver codepublic static void main(String[] args){    String str = "[|,[&,1,[!,0]],[!,[|,[|,1,0],[!,1]]]]";     System.out.println(logicalExpressionEvaluation(str));}} // This code is contributed by 29AjayKumar
# Python3 program to solve the# logical expression.import math as mt # Function to evaluate the logical expressiondef logicalExpressionEvaluation(string):     arr = list()     # traversing string from the end.    n = len(string)    for i in range(n - 1, -1, -1):        if (string[i] == "["):             s = list()             while (arr[-1] != "]"):                s.append(arr[-1])                arr.pop()             arr.pop()             # for NOT operation            if (len(s) == 3):                if s[2] == "1":                    arr.append("0")                else:                    arr.append("1")             # for AND and OR operation            elif (len(s) == 5):                a = int(s[0]) - 48                b = int(s[4]) - 48                c = 0                if s[2] == "&":                    c = a & b                else:                    c = a | b                arr.append((c) + 48)                     else:            arr.append(string[i])     return arr[-1] # Driver codestring= "[|,[&,1,[!,0]],[!,[|,[|,1,0],[!,1]]]]" print(logicalExpressionEvaluation(string)) # This code is contributed# by mohit kumar 29
// C# program to solve the logical expression.using System;using System.Collections.Generic; public class GFG{      // Function to evaluate the logical expressionstatic char logicalExpressionEvaluation(String str){    Stack<char> arr = new Stack<char>();      // traversing string from the end.    for (int i = str.Length - 1; i >= 0; i--)    {        if (str[i] == '[')        {            List<char> s = new List<char>();            while (arr.Peek() != ']')            {                s.Add(arr.Peek());                arr.Pop();            }            arr.Pop();              // for NOT operation            if (s.Count == 3)            {                arr.Push(s[2] == '1' ? '0' : '1');            }                          // for AND and OR operation            else if (s.Count == 5)            {                int a = s[0] - 48,                    b = s[4] - 48, c;                if(s[2] == '&' )                {                    c = a & b;                }                else                {                    c = a | b;                }                arr.Push((char)(c + 48));            }        }        else        {            arr.Push(str[i]);        }    }    return arr.Peek();}  // Driver codepublic static void Main(String[] args){    String str = "[[0,&,1],|,[!,1]]";      Console.WriteLine(logicalExpressionEvaluation(str));}} // This code is contributed by PrinciRaj1992
<script>// Javascript program to solve the logical expression. // Function to evaluate the logical expressionfunction logicalExpressionEvaluation(str){    let arr = [];      // traversing string from the end.    for (let i = str.length - 1; i >= 0; i--)    {        if (str[i] == '[')        {            let s = [];            while (arr[arr.length-1] != ']')            {                s.push(arr[arr.length-1]);                arr.pop();            }            arr.pop();              // for NOT operation            if (s.length == 3)            {                arr.push(s[2] == '1' ? '0' : '1');            }                          // for AND and OR operation            else if (s.length == 5)            {                let a = s[0].charCodeAt(0) - 48,                    b = s[4].charCodeAt(0) - 48, c;                                                          if(s[2] == '&' )                {                    c = a & b;                }                else                {                    c = a | b;                }                arr.push(String.fromCharCode(c + 48));            }        }        else        {            arr.push(str[i]);        }    }         return arr[arr.length-1];} // Driver codelet str = "[|,[&,1,[!,0]],[!,[|,[|,1,0],[!,1]]]]";document.write(logicalExpressionEvaluation(str));         // This code is contributed by patel2127</script>
0
Time Complexity: O(n) Here, n is the length of the string. 
Dhirendra121
mohit kumar 29
29AjayKumar
princiraj1992
ManishaYadav30
patel2127
programming-puzzle
Technical Scripter 2018
Stack
Strings
Technical Scripter
Strings
Stack
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