Compare Strings by Frequency of the Smallest Character

# 1170. Compare Strings by Frequency of the Smallest Character

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.
```

/**
 * @param {string[]} queries
 * @param {string[]} words
 * @return {number[]}
 */
 var numSmallerByFrequency = function(queries, words) {
    
    const getSmallCharFreq = (str) => {
        const obj = {};
        let smallChar = 'z';
        for (let i = 0; i < str.length; i++) {
            if (str[i] < smallChar) {
                smallChar = str[i];
            }
           obj[str[i]] = obj[str[i]] ? ++obj[str[i]] : 1;
        } 
        return obj[smallChar];
    };

     const qryFreq = [];
     for (let i = 0; i < queries.length; i++) {
        qryFreq.push(getSmallCharFreq(queries[i]));
     } 

     const wordFreq = [];
     for (let i = 0; i < words.length; i++) {
        wordFreq.push(getSmallCharFreq(words[i]));
     } 

    let out = [];
    for (let i = 0; i < qryFreq.length; i++) { 
         let cnt = 0;
         for (let j = 0; j < wordFreq.length; j++) {
             if (qryFreq[i] < wordFreq[j]) {
                cnt++;
             }
         }
         out.push(cnt); 
    }
    return out; 
};