Package g1101_1200.s1170_compare_strings_by_frequency_of_the_smallest_character

Class Solution

  • public class Solution
extends Object
    1170 - Compare Strings by Frequency of the Smallest Character.

    Medium

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

    You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words.

    Return an integer array answer, where each answer[i] is the answer to the ith query.

    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] consist of lowercase English letters.

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • numSmallerByFrequency

        public int[] numSmallerByFrequency​(String[] queries,
                                   String[] words)