1170. Compare Strings by Frequency of the Smallest Character
Description
Let the function f(s)
be the frequency of the lexicographically smallest character in a non-empty string s
. For example, if s = "dcce"
then f(s) = 2
because the lexicographically smallest character is 'c'
, which has a frequency of 2.
You are given an array of strings words
and another array of query strings queries
. For each query queries[i]
, count the number of words in words
such that f(queries[i])
< f(W)
for each W
in words
.
Return an integer array answer
, where each answer[i]
is the answer to the ith
query.
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]
consist of lowercase English letters.
Solutions
Solution 1: Sorting + Binary Search
First, according to the problem description, we implement a function $f(s)$, which returns the frequency of the smallest letter in the string $s$ in lexicographical order.
Next, we calculate $f(w)$ for each string $w$ in $words$, sort them, and store them in an array $nums$.
Then, we traverse each string $q$ in $queries$, and binary search in $nums$ for the first position $i$ that is greater than $f(q)$. Then, the elements at index $i$ and after in $nums$ all satisfy $f(q) < f(W)$, so the answer to the current query is $n - i$.
The time complexity is $O((n + q) \times M)$, and the space complexity is $O(n)$. Here, $n$ and $q$ are the lengths of the arrays $words$ and $queries$ respectively, and $M$ is the maximum length of the strings.
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