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

Minimum Distance Between Three Equal Elements I (LeetCode #3740) 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 hash map reduces the number of checks needed.

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

The optimal solution for Minimum Distance Between Three Equal Elements I runs in O(n) time and uses O(n) space. We traverse the array once to build the index map and then check each unique element's indices, leading to linear time complexity.

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

The brute force approach for Minimum Distance Between Three Equal Elements I is Brute Force, with O(n³) time complexity and O(1) space complexity. We can check all combinations of three indices to see if they point to the same value. This method is straightforward but inefficient as it checks every possible triplet. 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 I?

Minimum Distance Between Three Equal Elements I 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 I?

Explain your thought process clearly. Discuss edge cases like arrays with fewer than three elements. Practice similar problems to build confidence.

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

Not checking if there are at least three indices for an element. Confusing the distance formula with simple index differences.

#3740

Minimum Distance Between Three Equal Elements I

Easy

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n³)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a hash map to store the indices of each element. This allows us to quickly find the distances between indices of the same element without checking all combinations.

⚙️

Algorithm

3 steps

1Step 1: Create a hash map to store lists of indices for each unique element in nums.
2Step 2: For each element with at least three indices, calculate the distance using the first three indices.
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 = abs(indices[0] - indices[1]) + abs(indices[1] - indices[2]) + abs(indices[2] - indices[0])
10            min_dist = min(min_dist, dist)
11    return min_dist if min_dist != float('inf') else -1

ℹ

Complexity note: We traverse the array once to build the index map and then check each unique element's indices, leading to linear time complexity.

1Using a hash map reduces the number of checks needed.
2Focusing on indices of repeated elements allows for quick distance calculations.

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