How do you solve Longest Balanced Subarray II on LeetCode?

Longest Balanced Subarray II (LeetCode #3721) 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 hash maps allows efficient counting of distinct elements.

What is the time complexity of Longest Balanced Subarray II?

The optimal solution for Longest Balanced Subarray II runs in O(n) time and uses O(n) space. The algorithm efficiently counts distinct numbers using hash maps, leading to linear time complexity.

What is the brute force solution for Longest Balanced Subarray II?

The brute force approach for Longest Balanced Subarray II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check every possible subarray to see if it meets the balanced condition. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Longest Balanced Subarray II?

Explain your thought process clearly. Discuss edge cases and constraints upfront. Practice coding without an IDE to simulate real interview conditions.

What are common mistakes when solving Longest Balanced Subarray II?

Not resetting counts correctly when moving the left pointer. Overlooking the need to check for balance after each addition.

#3721

Longest Balanced Subarray II

Hard

ArrayHash TableDivide and ConquerSegment TreePrefix SumHash 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 track the counts of distinct even and odd numbers as we expand the subarray. Adjust the starting index when the counts are imbalanced.

⚙️

Algorithm

3 steps

1Step 1: Initialize two hash maps to count distinct even and odd numbers.
2Step 2: Use two pointers to expand the end of the subarray and adjust the start when the counts are unequal.
3Step 3: Update the maximum length whenever the counts are equal.

solution.py21 lines

1def longestBalancedSubarray(nums):
2    left = 0
3    even_count, odd_count = {}, {}
4    max_length = 0
5    for right in range(len(nums)):
6        if nums[right] % 2 == 0:
7            even_count[nums[right]] = even_count.get(nums[right], 0) + 1
8        else:
9            odd_count[nums[right]] = odd_count.get(nums[right], 0) + 1
10        while len(even_count) != len(odd_count):
11            if nums[left] % 2 == 0:
12                even_count[nums[left]] -= 1
13                if even_count[nums[left]] == 0:
14                    del even_count[nums[left]]
15            else:
16                odd_count[nums[left]] -= 1
17                if odd_count[nums[left]] == 0:
18                    del odd_count[nums[left]]
19            left += 1
20        max_length = max(max_length, right - left + 1)
21    return max_length

ℹ

Complexity note: The algorithm efficiently counts distinct numbers using hash maps, leading to linear time complexity.

1Using hash maps allows efficient counting of distinct elements.
2Two-pointer technique helps maintain balance dynamically.

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