Welcome to Subscribe On Youtube
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.
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