LC_Compare_String_by_Frequency_of_the_Smallest_Character.py

"""
1170. Compare Strings by Frequency of the Smallest Character

Let's define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = "dcce" then f(s) = 2 because the smallest character is "c" and its frequency is 2.

Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.

 

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
 

Constraints:

1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] are English lowercase letters.
"""

class Solution(object):
    def isPossibleDivide(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: bool
        """
        cc = Counter(nums)
        keys = [v for v in cc]
        keys.sort()
        
        idx = 0
        while idx < len(keys):
            start = keys[idx]
            for num in range(start, start+k):
                if cc[num] <= 0:
                    return False
                cc[num] -= 1
            
            while idx < len(keys) and cc[keys[idx]] == 0:
                idx += 1
        return True