1170

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.

Related Topics:
Array, String

Solution 1.

// OJ: https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
// Author: github.com/lzl124631x
// Time: O(W*Q)
// Space: O(W)
class Solution {
    int frequency(string &s) {
        int ch = 'z', cnt = 0;
        for (char c : s) {
            if (c < ch) {
                cnt = 1;
                ch = c;
            } else if (c == ch) ++cnt;
        }
        return cnt;
    }
public:
    vector<int> numSmallerByFrequency(vector<string>& Q, vector<string>& W) {
        vector<int> F(W.size()), ans;
        for (int i = 0; i < W.size(); ++i) F[i] = frequency(W[i]);
        for (auto &s : Q) {
            int cnt = 0, f = frequency(s);
            for (int i = 0; i < W.size(); ++i) cnt += F[i] > f;
            ans.push_back(cnt);
        }
        return ans;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
// Author: github.com/lzl124631x
// Time: O(W + Q)
// Space: O(1)
class Solution {
    int frequency(string &s) {
        int ch = 'z', cnt = 0;
        for (char c : s) {
            if (c < ch) {
                cnt = 1;
                ch = c;
            } else if (c == ch) ++cnt;
        }
        return cnt;
    }
public:
    vector<int> numSmallerByFrequency(vector<string>& Q, vector<string>& W) {
        vector<int> F(11), ans;
        for (int i = 0; i < W.size(); ++i) F[frequency(W[i])]++;
        for (int i = 1; i < 11; ++i) F[i] += F[i - 1];
        for (auto &s : Q) ans.push_back(W.size() - F[frequency(s)]);
        return ans;
    }
};