Solve the Logical Expression given by string

Given a 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]
0

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



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 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 string is fully traversed then return the value at stack top otherwise go to step 2.

Below is the implementation of the above approach:

C++

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// 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;
}

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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

Output:

0

Time Complexity : O(n) Here, n is length of string.



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