What is the time complexity of Find Maximum Number of Non Intersecting Substrings?

The optimal solution for Find Maximum Number of Non Intersecting Substrings runs in O(n) time and uses O(n) space. We traverse the string once, using a hash map for constant time lookups.

What is the brute force solution for Find Maximum Number of Non Intersecting Substrings?

The brute force approach for Find Maximum Number of Non Intersecting Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible substring of the given string to see if it meets the criteria. This method 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 Find Maximum Number of Non Intersecting Substrings?

Find Maximum Number of Non Intersecting Substrings uses the following concepts: Hash Table, String, Dynamic Programming, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Maximum Number of Non Intersecting Substrings?

Clarify the problem constraints. Discuss edge cases early. Think aloud to show your thought process.

What are common mistakes when solving Find Maximum Number of Non Intersecting Substrings?

Not checking for overlaps correctly. Ignoring substrings shorter than 4 characters.

#3557

Find Maximum Number of Non Intersecting Substrings

Medium

Hash TableStringDynamic ProgrammingGreedyHash 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)

We can track the starting and ending indices of valid substrings using a hash map. This allows us to efficiently count non-overlapping substrings.

⚙️

Algorithm

3 steps

1Step 1: Create a map to store the last occurrence of each character.
2Step 2: Iterate through the string, and for each character, check if a valid substring can be formed.
3Step 3: Count non-overlapping valid substrings by updating the last used index.

solution.py11 lines

1def maxNonIntersectingSubstrings(word):
2    last = {}
3    count = 0
4    end = -1
5    for i in range(len(word)):
6        if word[i] in last:
7            if i - last[word[i]] >= 4 and last[word[i]] > end:
8                count += 1
9                end = i
10        last[word[i]] = i
11    return count

ℹ

Complexity note: We traverse the string once, using a hash map for constant time lookups.

1Substrings must be at least 4 characters long.
2Non-overlapping substrings are crucial.

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