You are given an array of strings words and a string chars.
A string is good if it can be formed by characters from chars (each character can only be used once).
Return the sum of lengths of all good strings in words.
Example 1:
Input: words = ["cat","bt","hat","tree"], chars = "atach" Output: 6 Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.
Example 2:
Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr" Output: 10 Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.
Note:
1 <= words.length <= 10001 <= words[i].length, chars.length <= 100- All strings contain lowercase English letters only.
Solution: Hashtable
Use a hashtable to store each letter’s frequency of the string and compare that with each word.
Time complexity: O(n + sum(len(word))
Space complexity: O(1)
C++
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// Author: Huahua, 72 ms, 20 MB class Solution { public: int countCharacters(vector<string>& words, string chars) { vector<int> counts(26); for (char c : chars) ++counts[c - 'a']; int ans = 0; for (const string& word : words) { vector<int> cur(26); bool valid = true; for (char c: word) if (++cur[c - 'a'] > counts[c - 'a']) { valid = false; break; } if (valid) ans += word.length(); } return ans; } }; |
Python#
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# Author: Huahua class Solution: def countCharacters(self, words: List[str], chars: str) -> int: counts = collections.Counter(chars) def isValid(word): cur = collections.Counter(word) for c in cur: if c not in counts or cur[c] > counts[c]: return False return True ans = 0 for word in words: ans += len(word) if isValid(word) else 0 return ans |
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