How do you solve Mirror Frequency Distance on LeetCode?

Mirror Frequency Distance (LeetCode #3889) 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: Mirror characters are symmetric; understanding their mapping is key.

What is the time complexity of Mirror Frequency Distance?

The optimal solution for Mirror Frequency Distance runs in O(n) time and uses O(n) space. The frequency map allows for linear counting and retrieval, making the solution efficient.

What is the brute force solution for Mirror Frequency Distance?

The brute force approach for Mirror Frequency Distance is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks each character against all others to compute frequencies and their differences. It's straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Mirror Frequency Distance?

Mirror Frequency Distance uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Mirror Frequency Distance?

Explain your thought process clearly. Discuss edge cases, like empty strings or single characters. Practice coding under time constraints.

What are common mistakes when solving Mirror Frequency Distance?

Forgetting to handle characters that don't have mirrors. Not accounting for the frequency of mirror characters properly.

#3889

Mirror Frequency Distance

Medium

Hash TableStringCountingHash 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)

This approach uses a frequency map to count characters efficiently, allowing for quick access to their mirror frequencies.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map of all characters in the string.
2Step 2: For each unique character, compute its mirror and find the frequency difference using the map.
3Step 3: Sum all the absolute differences.

solution.py8 lines

1def mirror_frequency_distance(s):
2    from collections import Counter
3    freq = Counter(s)
4    total_diff = 0
5    for c in freq:
6        m = chr(219 - ord(c)) if c.isalpha() else chr(105 + ord(c))
7        total_diff += abs(freq[c] - freq[m])
8    return total_diff

ℹ

Complexity note: The frequency map allows for linear counting and retrieval, making the solution efficient.

1Mirror characters are symmetric; understanding their mapping is key.
2Using a frequency map optimizes counting operations.

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