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.

Solution:

class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        Map<String, Integer> map = new HashMap();
        for (String s : queries) {
            map.computeIfAbsent(s, k -> count(k));
        }
        for (String s : words) {
            map.computeIfAbsent(s, k -> count(k));
        }
        int[] res = new int[queries.length];
        for (int i = 0; i < queries.length; i ++) {
            for (int j = 0; j < words.length; j ++) {
                if (map.get(queries[i]) < map.get(words[j])) {
                    res[i] ++;
                }
            }
        }
        return res;
    }
    
    private int count(String s) {
        int[] freq = new int[26];
        for (int i = 0; i < s.length(); i ++) {
            freq[s.charAt(i) - 'a'] ++;
        }
        for (int i = 0; i < 26; i ++) {
            if (freq[i] > 0) {
                return freq[i];
            }
        }
        return 0;
    }
}