How do you solve Minimum Absolute Distance Between Mirror Pairs on LeetCode?

Minimum Absolute Distance Between Mirror Pairs (LeetCode #3761) 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 pairs depend on reversing numbers.

What is the time complexity of Minimum Absolute Distance Between Mirror Pairs?

The optimal solution for Minimum Absolute Distance Between Mirror Pairs runs in O(n) time and uses O(n) space. We traverse the array once and use a hash map for storage, leading to O(n) time and space.

What is the brute force solution for Minimum Absolute Distance Between Mirror Pairs?

The brute force approach for Minimum Absolute Distance Between Mirror Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of indices to see if they form a mirror pair. This is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Absolute Distance Between Mirror Pairs?

Minimum Absolute Distance Between Mirror Pairs uses the following concepts: Array, Hash Table, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Absolute Distance Between Mirror Pairs?

Explain your thought process clearly. Discuss edge cases like empty arrays. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Absolute Distance Between Mirror Pairs?

Not handling leading zeros after reversing. Forgetting to update the index map correctly.

#3761

Minimum Absolute Distance Between Mirror Pairs

Medium

ArrayHash TableMathHash 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 hash map to store the most recent index of each reversed number allows us to find mirror pairs efficiently in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Initialize a hash map to store the most recent index of reversed numbers.
2Step 2: Iterate through nums, reverse each number, and check if it exists in the map.
3Step 3: If it exists, calculate the distance and update the minimum if it's smaller.

solution.py9 lines

1def minAbsDist(nums):
2    index_map = {}
3    min_dist = float('inf')
4    for i, num in enumerate(nums):
5        rev_num = int(str(num)[::-1])
6        if rev_num in index_map:
7            min_dist = min(min_dist, i - index_map[rev_num])
8        index_map[num] = i
9    return min_dist if min_dist != float('inf') else -1

ℹ

Complexity note: We traverse the array once and use a hash map for storage, leading to O(n) time and space.

1Mirror pairs depend on reversing numbers.
2Using a hash map allows efficient lookups.

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