How do you solve Find the Longest Equal Subarray on LeetCode?

Find the Longest Equal Subarray (LeetCode #2831) 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 frequency map allows us to focus on each unique number, reducing unnecessary checks.

What is the brute force solution for Find the Longest Equal Subarray?

The brute force approach for Find the Longest Equal Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray and counting how many elements need to be deleted to make all elements equal. 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 Find the Longest Equal Subarray?

Find the Longest Equal Subarray uses the following concepts: Array, Hash Table, Binary Search, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Longest Equal Subarray?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach before writing code to ensure you are on the right track. Practice implementing both brute-force and optimal solutions to solidify your understanding.

What are common mistakes when solving Find the Longest Equal Subarray?

Not considering the frequency of elements when calculating deletions. Failing to reset the left pointer correctly in the sliding window approach.

#2831

Find the Longest Equal Subarray

Medium

ArrayHash TableBinary SearchSliding WindowHash 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)

The optimal solution uses a sliding window technique to efficiently find the longest subarray where we can delete up to k elements. This approach is faster because it avoids redundant checks and focuses on maintaining a valid window.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each number.
2Step 2: For each unique number in the array, maintain a sliding window of indices where this number appears.
3Step 3: Expand the window until the number of deletions needed exceeds k, then shrink the window from the left to maintain the condition.

solution.py20 lines

1# Full working Python code
2
3def longestEqualSubarray(nums, k):
4    from collections import defaultdict
5    freq_map = defaultdict(list)
6    for i, num in enumerate(nums):
7        freq_map[num].append(i)
8    max_length = 0
9    for indices in freq_map.values():
10        left = 0
11        for right in range(len(indices)):
12            deletions = indices[right] - indices[left] - right + left
13            while deletions > k:
14                left += 1
15                deletions = indices[right] - indices[left] - right + left
16            max_length = max(max_length, right - left + 1)
17    return max_length
18
19# Example usage
20print(longestEqualSubarray([1,3,2,3,1,3], 3))

ℹ

Complexity note: This complexity is due to the single pass through the array and maintaining a frequency map, making it much more efficient than the brute-force approach.

1Using a frequency map allows us to focus on each unique number, reducing unnecessary checks.
2The sliding window technique efficiently manages the range of indices while keeping track of deletions.

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