How do you solve Maximum Number of Occurrences of a Substring on LeetCode?

Maximum Number of Occurrences of a Substring (LeetCode #1297) 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: Sliding window technique is efficient for fixed-length substring problems.

What is the brute force solution for Maximum Number of Occurrences of a Substring?

The brute force approach for Maximum Number of Occurrences of a Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the given string and counting their occurrences. This method is straightforward but inefficient for larger strings due to its quadratic time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Occurrences of a Substring?

Maximum Number of Occurrences of a Substring 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 Number of Occurrences of a Substring?

Clarify constraints and edge cases before coding. Discuss your thought process and approach before jumping into code. Practice explaining your code as you write it to simulate interview conditions.

What are common mistakes when solving Maximum Number of Occurrences of a Substring?

Not considering overlapping substrings. Failing to update the frequency map correctly.

#1297

Maximum Number of Occurrences of a Substring

Medium

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)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window approach to efficiently count valid substrings of fixed length while maintaining a count of unique characters. This reduces the need to generate all substrings explicitly.

⚙️

Algorithm

5 steps

1Step 1: Use a sliding window of size minSize to traverse the string.
2Step 2: Maintain a frequency map of characters in the current window and a count of unique characters.
3Step 3: For each new character added to the window, update the frequency map and unique character count.
4Step 4: If the unique character count exceeds maxLetters, shrink the window from the left until it is valid again.
5Step 5: Count occurrences of valid substrings and keep track of the maximum count.

solution.py19 lines

1def maxFreq(s, maxLetters, minSize, maxSize):
2    from collections import defaultdict
3    count = defaultdict(int)
4    n = len(s)
5    freq = defaultdict(int)
6    unique_count = 0
7    left = 0
8    for right in range(n):
9        if freq[s[right]] == 0:
10            unique_count += 1
11        freq[s[right]] += 1
12        if right - left + 1 > minSize:
13            freq[s[left]] -= 1
14            if freq[s[left]] == 0:
15                unique_count -= 1
16            left += 1
17        if right - left + 1 == minSize and unique_count <= maxLetters:
18            count[s[left:right + 1]] += 1
19    return max(count.values(), default=0)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once with the sliding window technique. The space complexity is O(n) due to the frequency map storing characters and their counts.

1Sliding window technique is efficient for fixed-length substring problems.
2Maintaining a frequency map helps track character counts and unique characters.

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