Problem Description

In this problem, we are given two inputs: an array of strings called words and a single string called chars. Our task is to determine which strings in the words array are "good". A "good" string is one that can be completely formed using the characters in chars. Each character in chars may only be used once when forming a word. After identifying all the "good" strings, we must calculate the sum of their lengths and return that sum as the result.

For example, if words = ["cat", "bt", "hat", "tree"] and chars = "atach", only "cat" and "hat" are "good" because they can be formed using the characters in chars without reusing a single character. The lengths of "cat" and "hat" are 3 and 3, respectively, so the sum is 6. The goal of the problem is to implement a function that can do this calculation for any given words and chars.

Intuition

The solution approach can be divided into the following steps:

1. Count Characters in chars: We first count the frequency of each character in the string chars. This helps us know how many times we can use a particular character while forming words.

2. Iterate Over words: Next, we loop through each string in words and count the frequency of each character in the current string (w).

3. Check if a Word Can be Formed: To determine if a word is "good", we compare character frequencies of the current word with those in chars. If for each character in the word, the count in chars is greater than or equal to the count in the word, the word is "good".

4. Calculate and Sum Lengths: For each "good" string found, we add its length to a running total, ans.

5. Return the Result: Once all words have been checked, return the accumulated sum of lengths.

The beauty of this approach lies in the efficient use of character counting, which allows us to verify if a word can be formed without having to repeatedly search for characters in chars. By using a Python Counter object, which is essentially a specialized dictionary for counting hashable objects, we perform the needed comparisons succinctly and efficiently. The all function in Python is then used to ensure that all character counts in the word meet the availability requirement in chars, which is a very intuitive way to check the condition for every character simultaneously.

Solution Approach

The implementation of the solution uses a Counter from Python's collections module, which is a specialized dictionary designed for counting hashable objects. The Counter data structure is ideal for tracking the frequency of elements in an iterable.

Here is a step-by-step breakdown of how the solution is implemented:

1. Create a Counter for chars: First, a Counter is created for the string chars. This Counter object, named cnt, will provide a dictionary-like structure where each key is a character from chars and its corresponding value is the number of occurrences of that character.

   cnt = Counter(chars)

2. Initialize an Answer Variable: An integer ans is initialized to zero, which will hold the sum of lengths of all "good" strings in the array words.

   ans = 0

3. Loop Through Each Word in words: We iterate over each word w in the words array. For each word, a new Counter is created to count the occurrences of characters in that word.

   for w in words:
       wc = Counter(w)

4. Check if the Word is "Good": Using the all function, we check if every character c in the current word has a frequency count v that is less than or equal to its count in cnt. This ensures that each character required to form the word w is available in the quantity needed in chars.

   if all(cnt[c] >= v for c, v in wc.items()):
       ans += len(w)

   • If the condition is true for all characters, it means the word can be formed from the characters in chars, hence it is a "good" word. We add the length of the word to ans.
   • We do not modify the original cnt Counter because we do not want to affect the counts for the subsequent iteration of words. Instead, we just use its values to check the wc counts.

5. Return the Total Length: After iterating through all the words, the total length of all "good" words is stored in ans, which we return.

   return ans

This algorithm has a time complexity that is dependent on the total number of characters in all words (O(N) where N is the total number of characters) since we are counting characters for each word and iterating over each character count. The space complexity is O(U) where U is the total number of unique characters in chars and all words since Counter objects need to keep track of each unique character and its count.

Example Walkthrough

Consider a small example where we have words = ["hello", "world", "loop"] and chars = "hloeldorw".

1. Step 1: Count Characters in chars: We count the frequency of each character:

   h:1, l:2, o:2, e:1, d:1, r:1, w:1

   We have enough characters to potentially make the words "hello", "world", and "loop".

2. Step 2: Initialize an Answer Variable: We initialize ans to zero:

   ans = 0

3. Step 3: Iterate Over words: We iterate over each word:

   • For "hello":

     • Count characters: h:1, e:1, l:2, o:1
     • We check each character against our chars count and see that we can form "hello" with chars. Since "hello" is a "good" word, we add its length (5) to ans:

       ans += 5   # ans = 5

   • For "world":

     • Count characters: w:1, o:1, r:1, l:1, d:1
     • All characters are present in our chars count. "world" is also a "good" word, so we add its length (5) to ans:

       ans += 5   # ans = 10

   • For "loop":

     • Count characters: l:1, o:2, p:1
     • We have only 2 'o's and 1 'l' in chars, not enough to form the word "loop", so we do not add anything to ans for this word.

4. Step 4: Return the Total Length: After processing all the words, the value of ans is 10 (5+5). This is the sum of lengths of all "good" words. We return this as the final answer:

   return ans   # ans = 10

The sum of lengths of all "good" words that can be formed by the given chars is 10 in this example.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be analyzed as follows:

1. The creation of the cnt counter from the chars string: this operation takes O(m), where m is the length of the chars string since we must count the frequency of each character in chars.

2. The for-loop iterates over each word in the words list. Let the length of the list be n and the average length of the words be k.

3. Inside the loop, a new counter wc is created for each word: this operation also has a complexity of O(k).

4. The all function checks if all characters in w have a count less or equal to their count in cnt. This operation is O(k) as it checks each character's count (up to the total character count of a word).

Since steps 3 and 4 are within the loop iteration, they will run n times, which makes that part of the algorithm O(n*k).

Summing these up, the total time complexity is O(m + n*k).

Space Complexity

The space complexity can be analyzed as follows:

1. The cnt counter for chars utilizes O(u) space, where u is the unique number of characters in chars.

2. The wc counter for each word similarly utilizes O(v) space in the worst case, where v is the unique number of characters in that word.

3. However, since wc is constructed for one word at a time, O(v) space will be reused for each word, and v is bounded by the fixed size of the alphabet (u), so we can consider O(u) space for the counters.

Given that space used by variables like ans and temporary variables in the iteration is negligible relative to the size of the input, the overall space complexity is dominated by the space required for the counters, which is O(u) where u is the number of unique letters in chars and is bounded by the size of the character set used (e.g., the English alphabet, which has a fixed size of 26).

Therefore, the space complexity is O(u).

Learn more about how to find time and space complexity quickly using problem constraints.