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]); } //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())[0] ); } }