LeetCode: Compare Strings by Frequency of the Smallest Character Posted on August 25, 2019July 26, 2020 by braindenny Compare Strings by Frequency of the Smallest Character Similar Problems: CheatSheet: Leetcode For Code Interview CheatSheet: Common Code Problems & Follow-ups Tag: #array, #binarysearch 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. Github: code.dennyzhang.com Credits To: leetcode.com Leave me comments, if you have better ways to solve. Solution: // https://code.dennyzhang.com/compare-strings-by-frequency-of-the-smallest-character // Basic Ideas: array + binarysearch // Complexity: Time O((m+n)*log(m)), Space O(n) import "sort" func getFrequency(word string) int { chars := make([]int, 26) for _, ch := range word { chars[ch-'a']++ } for _, v := range chars { if v != 0 { return v } } return 0 } func numSmallerByFrequency(queries []string, words []string) []int { l := make([]int, len(words)) for i, word := range words { l[i] = getFrequency(word) } sort.Ints(l) res := make([]int, len(queries)) for i, query := range queries { count := getFrequency(query) left, right := 0, len(l)-1 // 3 // 1 2 3 3 3 4 4 5 // find the first bigger than the target for left <= right { mid := (right-left)/2+left if l[mid] <= count { left = mid+1 } else { right = mid-1 } res[i] = len(l)-left } } return res } Post Views: 0