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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;
    }
};

  • 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;
    }
};

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

        arr = [f(s) for s in words]
        arr.sort()
        n = len(arr)
        return [n - bisect_right(arr, f(q)) for q in queries]

############

# 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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