How do you solve Evaluate the Bracket Pairs of a String on LeetCode?

Evaluate the Bracket Pairs of a String (LeetCode #1807) 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: Using a HashMap allows for quick lookups of values associated with keys.

What is the time complexity of Evaluate the Bracket Pairs of a String?

The optimal solution for Evaluate the Bracket Pairs of a String runs in O(n) time and uses O(n) space. The complexity is O(n) because we only traverse the string once, and lookups in the HashMap are O(1).

What is the brute force solution for Evaluate the Bracket Pairs of a String?

The brute force approach for Evaluate the Bracket Pairs of a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves scanning the string for bracket pairs and replacing them one by one. This method is straightforward but inefficient, as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Evaluate the Bracket Pairs of a String?

Evaluate the Bracket Pairs of a String uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Evaluate the Bracket Pairs of a String?

Clearly explain your thought process and the approach you choose. Consider edge cases, such as empty strings or keys not in the knowledge. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Evaluate the Bracket Pairs of a String?

Not using a HashMap for fast key-value lookups. Failing to handle cases where keys are not found, leading to incorrect outputs.

#1807

Evaluate the Bracket Pairs of a String

Medium

ArrayHash TableStringHash MapArray

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 single pass through the string while leveraging a HashMap for quick lookups of keys. This reduces the time complexity significantly by avoiding nested loops.

⚙️

Algorithm

5 steps

1Step 1: Create a HashMap from the knowledge array for O(1) access to values.
2Step 2: Initialize an empty result string and a pointer to traverse the input string.
3Step 3: As you iterate through the string, check for '(' to identify keys.
4Step 4: When a key is found, extract it and look it up in the HashMap, appending the corresponding value or '?' to the result.
5Step 5: Continue until the end of the string, then return the result.

solution.py16 lines

1def evaluate(s, knowledge):
2    knowledge_map = {k: v for k, v in knowledge}
3    result = ''
4    i = 0
5    while i < len(s):
6        if s[i] == '(':  
7            j = i
8            while s[j] != ')':
9                j += 1
10            key = s[i+1:j]
11            result += knowledge_map.get(key, '?')
12            i = j + 1
13        else:
14            result += s[i]
15            i += 1
16    return result

ℹ

Complexity note: The complexity is O(n) because we only traverse the string once, and lookups in the HashMap are O(1).

1Using a HashMap allows for quick lookups of values associated with keys.
2Iterating through the string once is more efficient than nested iterations.

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