How do you solve Find Mirror Score of a String on LeetCode?

Find Mirror Score of a String (LeetCode #3412) 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 stack helps in efficiently finding mirror pairs.

What is the brute force solution for Find Mirror Score of a String?

The brute force approach for Find Mirror Score of a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each character in the string against all previous characters to find a mirror match. This is straightforward but inefficient as it requires checking every character multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Mirror Score of a String?

Find Mirror Score of a String uses the following concepts: Hash Table, String, Stack, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Mirror Score of a String?

Explain your thought process clearly while coding. Consider edge cases, like strings with no mirrors. Practice similar problems to recognize patterns.

What are common mistakes when solving Find Mirror Score of a String?

Not using a stack and checking all previous indices repeatedly. Confusing character mirror calculations.

#3412

Find Mirror Score of a String

Medium

Hash TableStringStackSimulationHash 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)

Using a stack allows us to efficiently track the indices of characters that can potentially form a mirror pair. This reduces the need to repeatedly check all previous characters.

⚙️

Algorithm

3 steps

1Step 1: Initialize a score variable to 0 and a stack to keep track of indices.
2Step 2: For each character at index i, check if the stack is not empty and if the character at the top of the stack is the mirror of s[i].
3Step 3: If a match is found, pop the index from the stack, mark both indices, and add the difference (i - j) to the score. Otherwise, push the current index onto the stack.

solution.py11 lines

1# Full working Python code
2score = 0
3s = 'aczzx'
4stack = []
5for i in range(len(s)):
6    if stack and s[stack[-1]] == chr(219 - ord(s[i])):
7        j = stack.pop()
8        score += i - j
9    else:
10        stack.append(i)
11print(score)

ℹ

Complexity note: The time complexity is O(n) because we process each character once. The space complexity is O(n) due to the stack that may store all indices in the worst case.

1Using a stack helps in efficiently finding mirror pairs.
2Understanding character encoding can simplify mirror calculations.

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