How do you solve Largest Substring Between Two Equal Characters on LeetCode?

Largest Substring Between Two Equal Characters (LeetCode #1624) 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 HashMap allows us to track indices efficiently.

What is the time complexity of Largest Substring Between Two Equal Characters?

The optimal solution for Largest Substring Between Two Equal Characters runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the string once. The space complexity is O(n) due to the HashMap storing the indices of characters.

What is the brute force solution for Largest Substring Between Two Equal Characters?

The brute force approach for Largest Substring Between Two Equal Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible pairs of characters in the string to find the longest substring between two equal characters. This is straightforward but inefficient, as it examines every combination. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Largest Substring Between Two Equal Characters?

Largest Substring Between Two Equal Characters uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Largest Substring Between Two Equal Characters?

Always clarify the problem statement and constraints. Think about edge cases, such as strings with no repeating characters. Explain your thought process clearly while coding.

What are common mistakes when solving Largest Substring Between Two Equal Characters?

Not considering the case where there are no repeating characters. Confusing substring length with index differences.

#1624

Largest Substring Between Two Equal Characters

Easy

Hash TableStringHash 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 uses a HashMap to store the first occurrence of each character. This allows us to find the maximum distance between the first and last occurrence of each character in a single pass, making it much more efficient.

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Algorithm

4 steps

1Step 1: Initialize a HashMap to store the first index of each character.
2Step 2: Iterate through the string, and for each character, check if it is already in the HashMap.
3Step 3: If it is, calculate the distance from its first occurrence to the current index and update max_length if this distance is greater.
4Step 4: If it is not in the HashMap, add it with its index.

solution.py9 lines

1def maxLengthBetweenEqualCharacters(s):
2    index_map = {}
3    max_length = -1
4    for i, char in enumerate(s):
5        if char in index_map:
6            max_length = max(max_length, i - index_map[char] - 1)
7        else:
8            index_map[char] = i
9    return max_length

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Complexity note: The time complexity is O(n) because we traverse the string once. The space complexity is O(n) due to the HashMap storing the indices of characters.

1Using a HashMap allows us to track indices efficiently.
2The longest substring can be found by calculating distances between first and last occurrences.

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