 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: 
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]);
}

//System.out.println(Arrays.toString(q_freq));
//System.out.println(Arrays.toString(w_freq));

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]){
counter++;
}
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);
}else{
freqMap.put(word.charAt(i)+"",1);
}
}
if(freqMap.size() == 1) return word.length();
return freqMap.get( (freqMap.keySet().toArray()) );
}
}