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Solve the Logical Expression given by string

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  • Difficulty Level : Medium
  • Last Updated : 21 Nov, 2022
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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:
Explanation:
[[0,&,1],|,[!,1]
[[0,&,1],|,0
[[0,&,1],|,0] 
[0,|,0] 
[0,|,0] 
[0
0

Input: str = “[!,[[0,&,[!,1]],|,[!,[[!,0],&,1]]]]” 
Output:

Approach: 

  1. Start traversing the string from the end.
  2. If [ found go to Step-3 otherwise push the characters into the stack.
    • Pop characters from the stack until the stack top becomes”]”. > 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.
  3. Perform the respective operations on the vector elements then push the result back into the stack.
  4. 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++




// C++ program to solve the logical expression.
#include <bits/stdc++.h>
using namespace std;
 
// Function to evaluate the logical expression
char 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 code
int main()
{
    string str = "[[0,&,1],|,[!,1]]";
 
    cout << logicalExpressionEvaluation(str) << endl;
 
    return 0;
}


Java




// Java program to solve the logical expression.
import java.util.*;
 
class GFG
{
     
// Function to evaluate the logical expression
static 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 code
public static void main(String[] args)
{
    String str = "[[0,&,1],|,[!,1]]";
 
    System.out.println(logicalExpressionEvaluation(str));
}
}
 
// This code is contributed by 29AjayKumar


Python3




# Python3 program to solve the
# logical expression.
import math as mt
 
# Function to evaluate the logical expression
def 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 code
string= "[[0,&,1],|,[!,1]]"
 
print(logicalExpressionEvaluation(string))
 
# This code is contributed
# by mohit kumar 29


C#




// C# program to solve the logical expression.
using System;
using System.Collections.Generic;
 
public class GFG
{
      
// Function to evaluate the logical expression
static 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 code
public static void Main(String[] args)
{
    String str = "[[0,&,1],|,[!,1]]";
  
    Console.WriteLine(logicalExpressionEvaluation(str));
}
}
 
// This code is contributed by PrinciRaj1992


Javascript




// C# program to solve the logical expression.
using System;
using System.Collections.Generic;
 
public class GFG
{
      
// Function to evaluate the logical expression
static 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 code
public static void Main(String[] args)
{
    String str = "[[0,&,1],|,[!,1]]";
  
    Console.WriteLine(logicalExpressionEvaluation(str));
}
}
 
// This code is contributed by PrinciRaj1992


Output

0

Complexity Analysis:

  • Time Complexity: O(n) Here, n is the length of the string.
  • Auxiliary Space: O(1)

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