How do you solve Maximum Substrings With Distinct Start on LeetCode?

Maximum Substrings With Distinct Start (LeetCode #3760) 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: Each distinct starting character can form a substring.

What is the time complexity of Maximum Substrings With Distinct Start?

The optimal solution for Maximum Substrings With Distinct Start runs in O(n) time and uses O(n) space. We traverse the string once, maintaining a map of last indices, leading to O(n) time and space complexity.

What is the brute force solution for Maximum Substrings With Distinct Start?

The brute force approach for Maximum Substrings With Distinct Start is Brute Force, with O(n²) time complexity and O(n²) space complexity. We can generate all possible substrings and check if they start with distinct characters. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Substrings With Distinct Start?

Maximum Substrings With Distinct Start uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Substrings With Distinct Start?

Explain your thought process clearly. Consider edge cases like single characters or repeated characters. Discuss both brute force and optimal approaches.

What are common mistakes when solving Maximum Substrings With Distinct Start?

Not updating the last occurrence correctly. Forgetting to check the end condition for substrings.

#3760

Maximum Substrings With Distinct Start

Medium

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n²)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can traverse the string while keeping track of the last occurrence of each character. This allows us to split the string efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array to track the last index of each character.
2Step 2: Traverse the string and determine the end of each substring based on the last occurrence of characters.
3Step 3: Count the number of distinct substrings by checking the end indices.

solution.py11 lines

1def maxDistinctSubstrings(s):
2    last_index = {c: -1 for c in set(s)}
3    end = 0
4    count = 0
5    for i, c in enumerate(s):
6        end = max(end, last_index[c])
7        if i == end:
8            count += 1
9            end = i + 1
10        last_index[c] = i
11    return count

ℹ

Complexity note: We traverse the string once, maintaining a map of last indices, leading to O(n) time and space complexity.

1Each distinct starting character can form a substring.
2The last occurrence of characters determines valid splits.

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