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This approach involves counting the frequency of each character in the 'chars' string. For each word, check if the word can be constructed using these characters, respecting their frequencies.
Time Complexity: O(N*K + C), where N = number of words, K = average length of the word, C = length of 'chars'.
Space Complexity: O(1), as the auxiliary space used is constant, i.e., the two arrays of size 26.
1def count_characters(words, chars):
2 from collections import Counter
3 char_count = Counter(chars)
4 total_length = 0
5 for word in words:
6 word_count = Counter(word)
7 for letter, cnt in word_count.items():
8 if cnt > char_count[letter]:
9 break
10 else:
11 total_length += len(word)
12 return total_length
13
14words = ["cat", "bt", "hat", "tree"]
15chars = "atach"
16print(count_characters(words, chars)) # Output: 6The Python solution utilizes the Counter class from the collections module to track the frequency of characters. It loops through each word, checking if the word can be composed using the available character counts and adds up the length if possible.
This approach utilizes hash maps to count the frequency of characters in both the 'chars' string and each word. The map allows for dynamic sizing and flexibility with character counts.
Time Complexity: O(N*K + C), where N = number of words, K = average length of the word, C = length of 'chars'.
Space Complexity: O(1), additional space overhead is minimal with the unordered maps.
1def count_characters(words, chars):
2 from
Using Python's Counter for character mapping, this solution compares each word's character needs to that available in 'chars'. It uses a single line check with all() to determine if the word can be formed.