LeetCode 1297 - Maximum Number of Occurrences of a Substring (Medium)

Given a string s, return the maximum number of ocurrences of any substring under the following rules:

• The number of unique characters in the substring must be less than or equal to maxLetters.
• The substring size must be between minSize and maxSize inclusive.

 

Example 1:

Input: s = "aababcaab", maxLetters = 2, minSize = 3, maxSize = 4
Output: 2
Explanation: Substring "aab" has 2 ocurrences in the original string.
It satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).

Example 2:

Input: s = "aaaa", maxLetters = 1, minSize = 3, maxSize = 3
Output: 2
Explanation: Substring "aaa" occur 2 times in the string. It can overlap.

Example 3:

Input: s = "aabcabcab", maxLetters = 2, minSize = 2, maxSize = 3
Output: 3

Example 4:

Input: s = "abcde", maxLetters = 2, minSize = 3, maxSize = 3
Output: 0

 

Constraints:

• 1 <= s.length <= 10^5
• 1 <= maxLetters <= 26
• 1 <= minSize <= maxSize <= min(26, s.length)
• s only contains lowercase English letters.

方法：暴力找出所有可能的substring

要注意最后的return要加上[0]。max需要至少两个数值。以及在count时候，取i的范围是range(len(s)-minSize+1)

time complexity：O(minSize*n) space：O(minSize*n)

class Solution:
    def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
        count = collections.Counter(s[i:i + minSize] for i in range(len(s) - minSize +1))
        print(count)
        return max([count[w] for w in count if len(set(w)) <= maxLetters] + [0])