Solve the Logical Expression given by string
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]
0
Input: 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.
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 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
// 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
# 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#
// 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 |
Output
0
Time Complexity: O(n) Here, n is the length of the string.
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