Problem Description

You are given a string s and three integer parameters: maxLetters, minSize, and maxSize. Your task is to find the maximum number of times any substring appears in s, where the substring must satisfy two conditions:

1. Character limit: The substring must contain at most maxLetters unique characters. For example, if maxLetters = 2, then substrings like "aa", "ab", "bb" are valid (having 1 or 2 unique characters), but "abc" is not valid (having 3 unique characters).

2. Length constraint: The length of the substring must be between minSize and maxSize (inclusive).

The goal is to find which valid substring appears the most frequently in s and return that frequency count.

For example, if s = "aababcaab", maxLetters = 2, minSize = 3, and maxSize = 4:

• The substring "aab" has 2 unique characters ('a' and 'b') and length 3, which satisfies both conditions
• "aab" appears 2 times in the string (at positions 0-2 and 6-8)
• This would be the maximum frequency among all valid substrings

The key insight in the solution is that if a longer substring meets the conditions and appears multiple times, then any of its substrings of length minSize will also meet the conditions and appear at least as many times. Therefore, we only need to check substrings of length exactly minSize to find the maximum frequency.

Intuition

The key observation is that if a substring of length k (where minSize ≤ k ≤ maxSize) appears n times in the string, then every substring of length minSize within that longer substring will also appear at least n times.

To understand why, consider a substring of length 5 that appears 3 times. Any substring of length 3 within those 5 characters will also appear at least 3 times (at the same positions where the length-5 substring appears). This means that checking longer substrings is redundant - we'll never find a longer substring that has a higher frequency than the best substring of length minSize.

Additionally, if a longer substring satisfies the unique character constraint (having at most maxLetters unique characters), then any of its substrings will also satisfy this constraint, since a substring cannot have more unique characters than the string containing it.

Therefore, instead of checking all possible substring lengths from minSize to maxSize, we can optimize by only checking substrings of exactly length minSize. This significantly reduces the search space while guaranteeing we find the maximum frequency.

The approach becomes straightforward:

1. Generate all substrings of length minSize
2. For each substring, check if it has at most maxLetters unique characters
3. Count the frequency of valid substrings using a hash table
4. Return the maximum frequency found

This reduces the time complexity from checking multiple lengths to just checking one optimal length, making the solution more efficient.

Learn more about Sliding Window patterns.

Solution Approach

The solution uses a hash table combined with a sliding window approach to efficiently find all substrings of length minSize and track their frequencies.

Step-by-step implementation:

1. Initialize data structures: We use a Counter (hash table) to store the frequency of each valid substring, and a variable ans to track the maximum frequency found.

2. Iterate through all possible starting positions: We loop through the string from index 0 to len(s) - minSize, ensuring we can always extract a substring of length minSize without going out of bounds.

3. Extract substring: For each position i, we extract a substring t = s[i : i + minSize] of exactly length minSize.

4. Check unique character constraint: We convert the substring t into a set ss = set(t) to get unique characters. The size of this set tells us how many unique characters are in the substring. If len(ss) <= maxLetters, the substring is valid.

5. Update frequency count: For valid substrings, we increment their count in the hash table using cnt[t] += 1.

6. Track maximum frequency: After updating the count, we immediately check if this is the new maximum frequency with ans = max(ans, cnt[t]).

7. Return result: After checking all possible substrings of length minSize, we return ans which contains the maximum frequency.

Time Complexity: O(n × minSize) where n is the length of string s. We iterate through n - minSize + 1 positions, and for each position, we create a substring of length minSize and convert it to a set.

Space Complexity: O(n × minSize) in the worst case, where all substrings are unique and stored in the hash table.

The elegance of this solution lies in recognizing that we only need to check substrings of the minimum allowed length, which significantly simplifies the problem while guaranteeing the optimal answer.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Input: s = "aabaab", maxLetters = 2, minSize = 2, maxSize = 3

Step 1: Initialize

• Create an empty hash table cnt = {}
• Set ans = 0 to track maximum frequency

Step 2: Extract all substrings of length minSize (2)

We'll iterate through positions 0 to 4 (since string length is 6 and we need substrings of length 2):

• Position 0: Extract "aa"

  • Unique characters: {'a'} → 1 unique character
  • 1 ≤ 2 (maxLetters), so it's valid
  • Update: cnt["aa"] = 1, ans = max(0, 1) = 1

• Position 1: Extract "ab"

  • Unique characters: {'a', 'b'} → 2 unique characters
  • 2 ≤ 2 (maxLetters), so it's valid
  • Update: cnt["ab"] = 1, ans = max(1, 1) = 1

• Position 2: Extract "ba"

  • Unique characters: {'b', 'a'} → 2 unique characters
  • 2 ≤ 2 (maxLetters), so it's valid
  • Update: cnt["ba"] = 1, ans = max(1, 1) = 1

• Position 3: Extract "aa"

  • Unique characters: {'a'} → 1 unique character
  • 1 ≤ 2 (maxLetters), so it's valid
  • Update: cnt["aa"] = 2, ans = max(1, 2) = 2

• Position 4: Extract "ab"

  • Unique characters: {'a', 'b'} → 2 unique characters
  • 2 ≤ 2 (maxLetters), so it's valid
  • Update: cnt["ab"] = 2, ans = max(2, 2) = 2

Step 3: Return result

• Final hash table: cnt = {"aa": 2, "ab": 2, "ba": 1}
• Maximum frequency: ans = 2

Result: The function returns 2, as both "aa" and "ab" appear twice in the string, which is the maximum frequency among all valid substrings.

Note how we ignored checking substrings of length 3 (maxSize). Even though "aab" appears twice and satisfies all constraints, we don't need to check it because any valid longer substring that appears n times will have at least one substring of minSize that also appears n times.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × m), where n is the length of the string s and m is minSize. This is because:

• The outer loop runs n - minSize + 1 times, which is O(n)
• For each iteration, we extract a substring of length minSize using slicing s[i : i + minSize], which takes O(m) time
• Creating a set from the substring takes O(m) time
• The Counter update and max comparison are O(1) operations
• Therefore, the total time complexity is O(n × m)

The space complexity is O(n × m), where n is the length of the string s and m is minSize. This is because:

• The Counter cnt can store at most n - minSize + 1 different substrings, which is O(n)
• Each substring stored as a key has length minSize, which is m
• Therefore, the total space needed to store all possible substrings is O(n × m)
• The temporary set ss uses O(m) space, but this doesn't affect the overall complexity

Note that in this problem, m (which is minSize) does not exceed 26 as per the reference answer.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Check All Substring Lengths from minSize to maxSize

One of the most common mistakes is trying to check all possible substring lengths between minSize and maxSize. This seems intuitive since the problem states the substring length should be "between minSize and maxSize", but it leads to:

• Unnecessary complexity: The code becomes more complex with nested loops
• Worse time complexity: O(n × (maxSize - minSize + 1) × maxSize) instead of O(n × minSize)
• Potential incorrect results: Without careful implementation, you might count overlapping substrings incorrectly

Incorrect approach:

def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
    max_frequency = 0
    substring_counter = Counter()
  
    # Wrong: Checking all lengths from minSize to maxSize
    for length in range(minSize, maxSize + 1):
        for start in range(len(s) - length + 1):
            substring = s[start : start + length]
            if len(set(substring)) <= maxLetters:
                substring_counter[substring] += 1
                max_frequency = max(max_frequency, substring_counter[substring])
  
    return max_frequency

Why the optimization works: If a substring of length k appears n times, then each of its substrings of length minSize will appear at least n times. Therefore, the maximum frequency can always be achieved by a substring of minimum length.

2. Incorrectly Calculating String Bounds

Another frequent error is miscalculating the loop boundaries, leading to index out of range errors:

Incorrect boundary:

# Wrong: This will cause index out of bounds
for start_index in range(len(s) - minSize):  # Missing +1
    current_substring = s[start_index : start_index + minSize]

Correct boundary:

# Correct: Ensures we can always extract minSize characters
for start_index in range(len(s) - minSize + 1):
    current_substring = s[start_index : start_index + minSize]

The +1 is crucial because when start_index = len(s) - minSize, we can still extract exactly minSize characters from position start_index to start_index + minSize - 1.

3. Using len() on the Substring Instead of Set for Unique Character Count

A subtle but critical mistake is checking the length of the substring instead of the number of unique characters:

Incorrect:

# Wrong: This checks substring length, not unique character count
if len(current_substring) <= maxLetters:
    substring_counter[current_substring] += 1

Correct:

# Correct: This checks the number of unique characters
if len(set(current_substring)) <= maxLetters:
    substring_counter[current_substring] += 1

4. Not Handling Edge Cases

Failing to consider edge cases can lead to unexpected results:

• Empty string or minSize > len(s): The loop won't execute, correctly returning 0
• maxLetters = 0: No valid substrings exist (unless considering empty substring, which isn't typically valid)
• minSize = 1: Every single character becomes a candidate

Solution: The provided code naturally handles these edge cases, but it's worth adding validation if needed:

if not s or minSize > len(s) or maxLetters <= 0:
    return 0