Package g1101_1200.s1170_compare_strings_by_frequency_of_the_smallest_character

Class Solution

java.lang.Object
g1101_1200.s1170_compare_strings_by_frequency_of_the_smallest_character.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 Details

    • Solution

      public Solution()

  • Method Details

    • numSmallerByFrequency

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