Leetcode 1170. Compare Strings by Frequency of the Smallest Character

Problem statement

https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/

Solution

For each word we evaluate frequeny of smallest symbol, then sort all these values in words. For each element from queries use binary search.

Complexity

It is O(n log n + m log n), where n = len(words) and m = len(queries), space is O(m + n).

Code

class Solution:
    def numSmallerByFrequency(self, queries, words):
        f = sorted(w.count(min(w)) for w in words)
        return [len(f) - bisect.bisect(f, q.count(min(q))) for q in queries]