How do you solve Longest Balanced Subarray I on LeetCode?

Longest Balanced Subarray I (LeetCode #3719) 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: Distinct counts of evens and odds must match.

What is the time complexity of Longest Balanced Subarray I?

The optimal solution for Longest Balanced Subarray I runs in O(n) time and uses O(n) space. We traverse the array once, leading to O(n) time. Space is O(n) due to the hashmap storing differences.

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

The brute force approach for Longest Balanced Subarray I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible subarray and count distinct even and odd numbers. If they match, we update the maximum length found. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Longest Balanced Subarray I?

Explain your thought process clearly. Consider edge cases, like all evens or odds. Discuss time and space complexity.

What are common mistakes when solving Longest Balanced Subarray I?

Not resetting counts for new subarrays. Confusing even/odd checks.

#3719

Longest Balanced Subarray I

Medium

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 hashmap to track the difference between counts of even and odd numbers, leveraging prefix sums to find balanced subarrays efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize a hashmap to store the first occurrence of each difference between even and odd counts.
2Step 2: Traverse the array, updating counts and calculating the difference.
3Step 3: If the difference has been seen before, calculate the length of the subarray and update the maximum length.

solution.py15 lines

1def longestBalanced(nums):
2    count_map = {0: -1}
3    even_count = odd_count = 0
4    max_length = 0
5    for i, num in enumerate(nums):
6        if num % 2 == 0:
7            even_count += 1
8        else:
9            odd_count += 1
10        diff = even_count - odd_count
11        if diff in count_map:
12            max_length = max(max_length, i - count_map[diff])
13        else:
14            count_map[diff] = i
15    return max_length

ℹ

Complexity note: We traverse the array once, leading to O(n) time. Space is O(n) due to the hashmap storing differences.

1Distinct counts of evens and odds must match.
2Using a hashmap allows efficient tracking of differences.

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