How do you solve Maximum Length Substring With Two Occurrences on LeetCode?

Maximum Length Substring With Two Occurrences (LeetCode #3090) 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 sliding window allows us to efficiently manage the substring without generating all possible substrings.

What is the brute force solution for Maximum Length Substring With Two Occurrences?

The brute force approach for Maximum Length Substring With Two Occurrences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible substrings of the given string to see if they meet the condition of having at most two occurrences of each character. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Length Substring With Two Occurrences?

Maximum Length Substring With Two Occurrences uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Maximum Length Substring With Two Occurrences?

Always think about how to optimize a brute-force solution, especially for string problems. Practice explaining your thought process clearly, as communication is key in interviews. Be prepared to discuss the trade-offs of different approaches, including time and space complexity.

What are common mistakes when solving Maximum Length Substring With Two Occurrences?

Failing to reset the left pointer correctly when the condition is violated. Not using a hashmap to count occurrences, leading to inefficient checks.

#3090

Maximum Length Substring With Two Occurrences

Easy

Hash TableStringSliding WindowHash MapSliding Window

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 solution uses a sliding window approach to maintain a substring that contains at most two occurrences of each character. This is efficient because it avoids checking all possible substrings and instead expands and contracts the window based on character counts.

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Algorithm

4 steps

1Step 1: Initialize two pointers (left and right) and a hashmap to count character occurrences.
2Step 2: Expand the right pointer to include characters until the condition is violated (more than two occurrences of any character).
3Step 3: When the condition is violated, move the left pointer to reduce the count of characters until the condition is satisfied again.
4Step 4: Keep track of the maximum length of valid substrings during the process.

solution.py13 lines

1def maxLengthSubstring(s):
2    left = 0
3    count = {}
4    max_length = 0
5    for right in range(len(s)):
6        count[s[right]] = count.get(s[right], 0) + 1
7        while count[s[right]] > 2:
8            count[s[left]] -= 1
9            if count[s[left]] == 0:
10                del count[s[left]]
11            left += 1
12        max_length = max(max_length, right - left + 1)
13    return max_length

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Complexity note: The time complexity is O(n) because each character is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(n) due to the hashmap storing character counts.

1Using a sliding window allows us to efficiently manage the substring without generating all possible substrings.
2Maintaining a character count helps in quickly determining if the substring meets the condition.

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