1160 Find Words That Can Be Formed By Characters - Easy

Problem:

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.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length, chars.length <= 100
• words[i] and chars consist of lowercase English letters.

Problem Analysis:

• Create a hashmap (count) to store the count of each character in the chars string.
• Iterate through a nested loop for each word in the list of words and each character in each word:
  • Check if the character exists in the count hashmap.
  • Verify that the count of the character in the current word is less than or equal to its count in the hashmap.
• Sum the lengths of all "good" words to obtain the final output.

Time Complexity:

• O(N * K + M), where N is the average length of words in the list words, K is the number of words in words, and M is the length of the chars string.

Space Complexity:

• O(N + M), where N represents the space required for the hashmap (count), and M is the size of the character set in chars.

Solutions:

class Solution:
    def countCharacters(self, words: List[str], chars: str) -> int:
        count = Counter(chars)
        res = 0
        

        for word in words:
            cur_word = defaultdict(int)

            good = True
            for char in word:
                cur_word[char] += 1
                if char not in count or cur_word[char] > count[char]:
                    # char dont exist or 
                    # curr word has more char than the counter
                    good = False
                    break
            if good:
                res += len(word)

        return res

Similar Questions