How do you solve Minimum Distance Between Three Equal Elements II on LeetCode?

Minimum Distance Between Three Equal Elements II (LeetCode #3741) 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: A good tuple requires at least three occurrences of the same element.

What is the time complexity of Minimum Distance Between Three Equal Elements II?

The optimal solution for Minimum Distance Between Three Equal Elements II runs in O(n) time and uses O(n) space. The complexity is linear due to a single pass to build the index map and another pass to calculate distances.

What is the brute force solution for Minimum Distance Between Three Equal Elements II?

The brute force approach for Minimum Distance Between Three Equal Elements II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all combinations of three indices to find good tuples. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Distance Between Three Equal Elements II?

Minimum Distance Between Three Equal Elements II uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Distance Between Three Equal Elements II?

Explain your thought process clearly. Consider edge cases like small arrays. Discuss time and space complexity during your solution.

What are common mistakes when solving Minimum Distance Between Three Equal Elements II?

Not checking if the number appears at least three times. Miscalculating the distance using incorrect indices.

#3741

Minimum Distance Between Three Equal Elements II

Medium

ArrayHash TableHash 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)

Use a hash map to store indices of each number, allowing efficient retrieval of positions to calculate distances.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store indices of each number in the array.
2Step 2: For each number appearing at least 3 times, calculate the distance using the formula 2 * (max(i, j, k) - min(i, j, k)).
3Step 3: Return the minimum distance found, or -1 if no valid tuples exist.

solution.py11 lines

1def minDistance(nums):
2    from collections import defaultdict
3    index_map = defaultdict(list)
4    for i, num in enumerate(nums):
5        index_map[num].append(i)
6    min_dist = float('inf')
7    for indices in index_map.values():
8        if len(indices) >= 3:
9            dist = 2 * (indices[-1] - indices[0])
10            min_dist = min(min_dist, dist)
11    return min_dist if min_dist != float('inf') else -1

ℹ

Complexity note: The complexity is linear due to a single pass to build the index map and another pass to calculate distances.

1A good tuple requires at least three occurrences of the same element.
2Distance can be simplified to focus on the maximum and minimum indices.

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