How do you solve Parsing A Boolean Expression on LeetCode?

Parsing A Boolean Expression (LeetCode #1106) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding operator precedence is crucial.

What is the brute force solution for Parsing A Boolean Expression?

The brute force approach for Parsing A Boolean Expression is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves recursively parsing the expression and evaluating it as we encounter different components. We will handle each type of expression separately, which can lead to repeated evaluations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Parsing A Boolean Expression?

Parsing A Boolean Expression uses the following concepts: String, Stack, Recursion. The recommended patterns to study are: Stack, Recursion.

What are the interview tips for Parsing A Boolean Expression?

Clarify the input format and constraints. Think aloud while coding to demonstrate your thought process. Test your solution with edge cases.

What are common mistakes when solving Parsing A Boolean Expression?

Not handling parentheses correctly. Forgetting to evaluate the logical operations properly.

#1106

Parsing A Boolean Expression

Hard

StringStackRecursionStackRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to manage the parsing of the expression, allowing us to evaluate it in a single pass. This avoids redundant evaluations and keeps track of the depth of nested expressions efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize a stack to keep track of expressions and their results.
2Step 2: Iterate through the expression character by character, pushing operators and operands onto the stack.
3Step 3: When encountering a closing parenthesis, pop from the stack until the corresponding opening parenthesis is found, evaluate the expression, and push the result back onto the stack.
4Step 4: At the end of the iteration, the stack will contain the final result.

solution.py24 lines

1def parse(expression):
2    stack = []
3    i = 0
4    while i < len(expression):
5        if expression[i] in 'tf':
6            stack.append(expression[i] == 't')
7        elif expression[i] == '(':  
8            stack.append(i)
9        elif expression[i] == ')':
10            inner = []
11            while stack and isinstance(stack[-1], bool):
12                inner.append(stack.pop())
13            inner.reverse()
14            operator = stack.pop()
15            if operator == '!':
16                stack.append(not inner[0])
17            elif operator == '&':
18                stack.append(all(inner))
19            elif operator == '|':
20                stack.append(any(inner))
21        else:
22            stack.append(expression[i])
23        i += 1
24    return stack[0]

ℹ

Complexity note: This complexity is due to the single pass through the expression and the use of a stack to store intermediate results, which can grow with the depth of nested expressions.

1Understanding operator precedence is crucial.
2Recognizing the structure of nested expressions helps in parsing.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.