# 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.

• class Solution {
public int[] numSmallerByFrequency(String[] queries, String[] words) {
int n = words.length;
int[] nums = new int[n];
for (int i = 0; i < n; ++i) {
nums[i] = f(words[i]);
}
Arrays.sort(nums);
int m = queries.length;
int[] ans = new int[m];
for (int i = 0; i < m; ++i) {
int x = f(queries[i]);
int l = 0, r = n;
while (l < r) {
int mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
ans[i] = n - l;
}
return ans;
}

private int f(String s) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
++cnt[s.charAt(i) - 'a'];
}
for (int x : cnt) {
if (x > 0) {
return x;
}
}
return 0;
}
}

• class Solution {
public:
vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {
auto f = [](string s) {
int cnt[26] = {0};
for (char c : s) {
cnt[c - 'a']++;
}
for (int x : cnt) {
if (x) {
return x;
}
}
return 0;
};
int n = words.size();
int nums[n];
for (int i = 0; i < n; i++) {
nums[i] = f(words[i]);
}
sort(nums, nums + n);
vector<int> ans;
for (auto& q : queries) {
int x = f(q);
ans.push_back(n - (upper_bound(nums, nums + n, x) - nums));
}
return ans;
}
};

• class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
def f(s: str) -> int:
cnt = Counter(s)
return next(cnt[c] for c in ascii_lowercase if cnt[c])

n = len(words)
nums = sorted(f(w) for w in words)
return [n - bisect_right(nums, f(q)) for q in queries]


• func numSmallerByFrequency(queries []string, words []string) (ans []int) {
f := func(s string) int {
cnt := [26]int{}
for _, c := range s {
cnt[c-'a']++
}
for _, x := range cnt {
if x > 0 {
return x
}
}
return 0
}
n := len(words)
nums := make([]int, n)
for i, w := range words {
nums[i] = f(w)
}
sort.Ints(nums)
for _, q := range queries {
x := f(q)
ans = append(ans, n-sort.SearchInts(nums, x+1))
}
return
}

• function numSmallerByFrequency(queries: string[], words: string[]): number[] {
const f = (s: string): number => {
const cnt = new Array(26).fill(0);
for (const c of s) {
cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
}
return cnt.find(x => x > 0);
};
const nums = words.map(f).sort((a, b) => a - b);
const ans: number[] = [];
for (const q of queries) {
const x = f(q);
let l = 0,
r = nums.length;
while (l < r) {
const mid = (l + r) >> 1;
if (nums[mid] > x) {
r = mid;
} else {
l = mid + 1;
}
}
ans.push(nums.length - l);
}
return ans;
}