# 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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to solve the logical expression. ` `#include ` `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 == ``'1'` `? arr.push(``'0'``) : arr.push(``'1'``); ` `            ``} ` `            ``// for AND and OR operation ` `            ``else` `if` `(s.size() == 5) ` `            ``{ ` `                ``int` `a = s - 48, b = s - 48, c; ` `                ``s == ``'&'` `? 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 arr = ``new` `Stack(); ` ` `  `    ``// traversing string from the end. ` `    ``for` `(``int` `i = str.length() - ``1``; i >= ``0``; i--)  ` `    ``{ ` `        ``if` `(str.charAt(i) == ``'['``)  ` `        ``{ ` `            ``Vector s = ``new` `Stack(); ` `            ``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 == ``'1'` `? ``'0'` `: ``'1'``); ` `            ``} ` `              `  `            ``// for AND and OR operation ` `            ``else` `if` `(s.Count == 5) ` `            ``{ ` `                ``int` `a = s - 48,  ` `                    ``b = s - 48, c; ` `                ``if``(s == ``'&'` `) ` `                ``{ ` `                    ``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
```

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

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