LeetCode - Strings - 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").
class Solution {
    public int[] numSmallerByFrequency(String[] queries, String[] words) {
        int[] q_freq = new int[queries.length];
        for(int i=0;i<queries.length;i++){
            q_freq[i] = getFrequency(queries[i]);
        int[] w_freq = new int[words.length];
        for(int i=0;i<words.length;i++){
            w_freq[i] = getFrequency(words[i]);
        int[] result_freq = new int[queries.length];
        for(int i=0;i<q_freq.length;i++){
            int queryfrequency = q_freq[i];
            int counter = 0;
            for(int j=0;j<w_freq.length;j++){
                if(queryfrequency < w_freq[j]){
                result_freq[i] = counter;
        return result_freq;
    public int getFrequency(String word){
        if(word.length() == 1) return 1;
        Map<String,Integer> freqMap = new TreeMap<String,Integer>();
        for(int i=0;i<word.length();i++){
            if(freqMap.containsKey(word.charAt(i)+"")){                                     freqMap.put(word.charAt(i)+"",freqMap.get(word.charAt(i)+"")+1);
        if(freqMap.size() == 1) return word.length();
        return freqMap.get( (freqMap.keySet().toArray())[0] );

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