How do you solve Maximum Number of Non-Overlapping Substrings on LeetCode?

Maximum Number of Non-Overlapping Substrings (LeetCode #1520) 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: Understanding character occurrences is crucial for substring selection.

What is the brute force solution for Maximum Number of Non-Overlapping Substrings?

The brute force approach for Maximum Number of Non-Overlapping Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings and checking if they meet the criteria. This is straightforward but inefficient, as it examines every combination of substrings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Non-Overlapping Substrings?

Maximum Number of Non-Overlapping Substrings uses the following concepts: Hash Table, String, Greedy, Sorting. The recommended patterns to study are: Hash Map, Greedy Algorithm.

What are the interview tips for Maximum Number of Non-Overlapping Substrings?

Clearly explain your thought process and how you are approaching the problem. Consider edge cases, such as strings with all unique characters or repeated characters. Practice explaining your solution as you code to simulate the interview environment.

What are common mistakes when solving Maximum Number of Non-Overlapping Substrings?

Failing to account for all occurrences of a character in a substring. Not checking for overlaps correctly when selecting substrings.

#1520

Maximum Number of Non-Overlapping Substrings

Hard

Hash TableStringGreedySortingHash MapGreedy Algorithm

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 involves using a greedy approach with a two-pass technique. We first determine the last occurrence of each character, then we find the maximum number of non-overlapping substrings by iterating through the string and ensuring we include all occurrences of each character.

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Algorithm

3 steps

1Step 1: Create a map to store the last occurrence index of each character.
2Step 2: Iterate through the string to determine the end index for each substring based on the last occurrence of characters.
3Step 3: Use a greedy approach to select non-overlapping substrings by checking if the current substring can be added without overlap.

solution.py12 lines

1def maxNonOverlappingSubstrings(s):
2    last = {c: i for i, c in enumerate(s)}
3    result = []
4    end = -1
5    start = 0
6    while start < len(s):
7        current_end = last[s[start]]
8        for i in range(start, current_end + 1):
9            current_end = max(current_end, last[s[i]])
10        result.append(s[start:current_end + 1])
11        start = current_end + 1
12    return result

ℹ

Complexity note: The time complexity is O(n) because we make a single pass to determine last occurrences and another pass to find substrings. The space complexity is O(n) due to the storage of last occurrences in a map.

1Understanding character occurrences is crucial for substring selection.
2Greedy algorithms can effectively maximize non-overlapping selections.

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