How do you solve Minimum Seconds to Equalize a Circular Array on LeetCode?

Minimum Seconds to Equalize a Circular Array (LeetCode #2808) 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: Understanding the circular nature of the array helps in calculating gaps correctly.

What is the brute force solution for Minimum Seconds to Equalize a Circular Array?

The brute force approach for Minimum Seconds to Equalize a Circular Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying to equalize the array by checking every possible target value and calculating how many seconds it would take to make all elements equal to that value. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Seconds to Equalize a Circular Array?

Minimum Seconds to Equalize a Circular Array uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Seconds to Equalize a Circular Array?

Always clarify the problem constraints and edge cases before jumping into coding. Think about the time complexity of your solution and how it can be optimized. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Seconds to Equalize a Circular Array?

Not considering the circular nature of the array when calculating gaps. Overlooking the need to handle cases where the same number appears multiple times.

#2808

Minimum Seconds to Equalize a Circular Array

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach leverages the observation that we can minimize the number of seconds by focusing on the segments between occurrences of the same value. We only need to calculate the maximum distance between consecutive occurrences of the same value.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to store the indices of each unique value in the array.
2Step 2: For each unique value, calculate the maximum gap between consecutive indices to determine how many seconds are needed to fill the gaps.
3Step 3: Return the maximum of these values as the result.

solution.py16 lines

1# Full working Python code
2from collections import defaultdict
3
4def min_seconds(nums):
5    index_map = defaultdict(list)
6    for i, num in enumerate(nums):
7        index_map[num].append(i)
8    min_seconds = 0
9    n = len(nums)
10    for indices in index_map.values():
11        max_gap = 0
12        for j in range(1, len(indices)):
13            max_gap = max(max_gap, (indices[j] - indices[j - 1]) // 2)
14        max_gap = max(max_gap, (n - indices[-1] + indices[0]) // 2)
15        min_seconds = max(min_seconds, max_gap)
16    return min_seconds

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the array to build the index map and then another pass through the unique values, which is efficient given the constraints.

1Understanding the circular nature of the array helps in calculating gaps correctly.
2Using a map to store indices allows us to efficiently calculate the required seconds.

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