LeetCode [1170] Compare Strings by Frequency of the Smallest Character

 1170. Compare Strings by Frequency of the Smallest Character

Easy

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.

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class Solution { int getFreq(String word) { int[] freq = new int[26]; for(char c:word.toCharArray()) { freq[c-'a']++; } for(int i=0; i<26; ++i) { if(freq[i]!=0) return freq[i]; } return 0; } public int[] numSmallerByFrequency(String[] queries, String[] words) { int[] qF = new int[queries.length]; for(int i=0; i<queries.length; ++i) qF[i] = getFreq(queries[i]); int[] wF = new int[words.length]; for(int i=0; i<words.length; ++i) wF[i] = getFreq(words[i]); Arrays.sort(wF); int[] ret = new int[queries.length]; for(int i=0; i<queries.length; ++i){ int l = 0, r = wF.length-1; //find l as the first one with wF[l]>qF[i]; //l will also be the lenth of words whose frequence<=qF[i] while(l<=r) { int m = l + (r-l)/2; if(wF[m]<=qF[i]) { l = m+1; }else{ r = m-1; } } ret[i] = words.length - l; } return ret; } }