Formatted question description: https://leetcode.ca/all/1170.html

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/

// 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/

// 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;
    }
};

Java

  • class Solution {
        public int[] numSmallerByFrequency(String[] queries, String[] words) {
            int queriesLength = queries.length, wordsLength = words.length;
            int[] queriesFrequency = new int[queriesLength];
            int[] wordsFrequency = new int[wordsLength];
            for (int i = 0; i < queriesLength; i++)
                queriesFrequency[i] = frequency(queries[i]);
            for (int i = 0; i < wordsLength; i++)
                wordsFrequency[i] = frequency(words[i]);
            int[] nums = new int[queriesLength];
            for (int i = 0; i < queriesLength; i++) {
                int queryFrequency = queriesFrequency[i];
                for (int j = 0; j < wordsLength; j++) {
                    int wordFrequency = wordsFrequency[j];
                    if (queryFrequency < wordFrequency)
                        nums[i]++;
                }
            }
            return nums;
        }
    
        public int frequency(String str) {
            int[] count = new int[26];
            char[] array = str.toCharArray();
            for (char c : array)
                count[c - 'a']++;
            for (int i = 0; i < 26; i++) {
                if (count[i] != 0)
                    return count[i];
            }
            return 0;
        }
    }
    
  • // OJ: https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
    // 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;
        }
    };
    
  • # 1170. Compare Strings by Frequency of the Smallest Character
    # https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/
    
    class Solution:
        def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
            res, res1, res2 = [], [], []
            
            def f(word):
                count = [0] * 26
                for w in word:
                    count[ord(w) - ord('a')] += 1
    
                for c in count:
                    if c != 0: return c
                
                return -1
                
            for word in queries:
                res1.append(f(word))
            
            for word in words:
                res2.append(f(word))
            
            for x in res1:
                count = 0
                for y in res2:
                    if y > x:
                        count += 1
                
                res.append(count)
            
            return res
            
            
                
    
    

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