Problem Description

You are provided with two lists of strings: startWords and targetWords. Each string is composed of lowercase English letters. Your task is to determine how many strings in targetWords can be formed from any string in startWords by performing a specific conversion operation.

The conversion operation consists of two steps:

1. Append a single lowercase letter that is not already in the string to its end. For example, if you have the string "abc", you can add "d", "e", "y", but not "a" to it, creating strings like "abcd".
2. Rearrange the letters of the newly formed string in any order. For instance, the string "abcd" could be rearranged into "acbd", "bacd", "cbda", etc.

You need to count the number of strings in targetWords that can be achieved by applying this operation on any of the strings in startWords.

It's important to note that startWords aren't actually altered during this process; the operation is only used to verify the possibility of transforming a startWord into a targetWord.

Intuition

The solution is based on a clever use of bitwise operations to track the presence of each letter in a given string. We use an integer (bitmask) to represent each string, where the ith bit (from right to left) is set to 1 if the letter ‘a’ + i is in the string. For instance, the string "abc" would result in the bitmask 0b111 (in binary), representing the presence of 'a', 'b', and 'c'.

By using this approach, we can easily check whether we can form a targetWord from any startWord by appending a letter and rearranging it. For each targetWord, we can generate its bitmask and then, for each letter in that targetWord, we toggle the corresponding bit (using XOR operation) to simulate the removal of the letter. We then check if the resulting bitmask is present in the set of bitmasks created from the startWords. If it's found, we know that we can form the targetWord from that startWord.

Let’s walk through the steps of the solution:

1. Create an integer set s. For each word in startWords, calculate a bitmask representing the letters in the word and add it to s.
2. Initialize a counter ans to zero, which will keep track of the number of targetWords that can be formed.
3. For every word in targetWords, compute the bitmask in a similar way.
4. For each letter in the current targetWord, create a temporary bitmask by toggling (using XOR operation) the bit corresponding to the current letter. If the resulting bitmask is found in s, increment ans by one, and proceed to check the next targetWord.
5. Return the final count ans.

By representing strings as bitmasks, we switch from an O(N*26) character comparison problem to an O(N) integer comparison problem, which is much more efficient.

Learn more about Sorting patterns.

Solution Approach

The solution implements a bit manipulation strategy using Python's bitwise operations to encode each word as a bitmask, where each bit represents the presence of a corresponding letter from the alphabet.

Let's detail the steps followed in the given Python code:

1. Bitmask Creation for startWords:

   • Initialize an empty set s which will hold the unique bitmasks of all the words in startWords.
   • Iterate over each word in startWords, and for each word:
     • Initialize a variable mask with a value of 0. This mask will be the bitmask representation of the word.
     • For each character c in the word, calculate the difference ord(c) - ord('a') to find the position of the bit that corresponds to the character in the alphabet (0 for 'a', 1 for 'b', and so on).
     • Use the bitwise OR mask |= 1 << (ord(c) - ord('a')) operation to set the bit at the calculated position. This ensures that each bit in the mask represents whether a particular character is in the word.
   • Add the resulting mask to the set s.

2. Verification for targetWords:

   • Initialize the ans counter to 0, which will count the valid targetWords.
   • Repeat the similar process for each word in targetWords to create the bitmask.
   • For each character c in the current targetWord, generate a temporary bitmask t by toggling the bit corresponding to c in the word's bitmask. Toggling is done using the XOR operation mask ^ (1 << (ord(c) - ord('a'))), which effectively simulates removing the character c from the word.
   • If the temporary bitmask t is found in the set s, increment the ans because this indicates that the original targetWord can be formed by adding c to some startWord and rearranging the letters.
   • If a match is found in the set s for the current targetWord, break the inner loop to prevent counting the same targetWord multiple times, and continue with the next targetWord.

3. Return the Result:

   • After iterating over all the words in targetWords, return the final count ans.

The algorithm leverages bitwise operations for efficient comparison and the set data structure for constant-time lookup to check the possibility of formation of targetWords. The key patterns used here are bitwise encoding and set membership checks, which make the solution compact and efficient.

Example Walkthrough

To elaborate on the solution approach, let's walk through a small example:

Let's say we have:

• startWords = ["go","bat"]
• targetWords = ["bago","atb","tabg"]

Now, we will apply the solution approach on this example.

Bitmask Creation for startWords:

1. Initialize an empty set s. This will store the bitmasks of startWords.
2. For "go":
   • Start with bitmask mask = 0.
   • We add 'g' (bitmask 1 << (ord('g') - ord('a')) = 1 << 6).
   • We add 'o' (bitmask 1 << (ord('o') - ord('a')) = 1 << 14).
   • Final bitmask for "go" is 0 | 1 << 6 | 1 << 14, which is 0b1000001000000.
   • Add this bitmask to set s.
3. For "bat":
   • Start with bitmask mask = 0.
   • We add 'b' (bitmask 1 << (ord('b') - ord('a')) = 1 << 1).
   • We add 'a' (bitmask 1 << (ord('a') - ord('a')) = 1 << 0).
   • We add 't' (bitmask 1 << (ord('t') - ord('a')) = 1 << 19).
   • Final bitmask for "bat" is 0 | 1 << 1 | 1 << 0 | 1 << 19, which is 0b1000000000011.
   • Add this bitmask to set s.

After processing startWords, our set s has bitmasks {0b1000001000000, 0b1000000000011}.

Verification for targetWords:

1. Initialize ans to 0.
2. Process targetWords similarly and compare with bitmasks in s.
3. For "bago":
   • Bitmask for "bago" is 0b1010001000010.
   • Check each letter by toggling it in the bitmask:
     • Toggle 'b': We get mask 0b1010001000010 ^ 1 << 1 = 0b1010001000000 which is not in s.
     • Toggle 'a': We get mask 0b1010001000010 ^ 1 << 0 = 0b1010001000011 which is in s.
     • No need to check further, increment ans to 1.
4. For "atb":
   • Bitmask for "atb" directly matches the bitmask of "bat" in s.
   • Since one letter has to be added, this shouldn't have happened; "atb" is the same length as "bat", thus, we skip and don't increment ans.
5. For "tabg":
   • Bitmask for "tabg" is 0b1000001000011.
   • Check each letter by toggling it in the bitmask:
     • Toggle 't': We get mask 0b1000001000011 ^ 1 << 19 = 0b0000001000011, which matches "ab" (not in s, and too short anyway).
     • Toggle 'a': We get mask 0b1000001000011 ^ 1 << 0 = 0b1000001000010, which is in s.
     • No need to check further, increment ans to 2.

Return the Result:

After examining all targetWords, the final ans is 2, since the words "bago" and "tabg" in targetWords can be formed from the startWords "go" and "bat", respectively.

The key insights from this example that lead to the count of valid targetWords are the efficient bitmask representation of the words and constant-time set membership checks to verify the transformation possibilities.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(L * (N + M)), where L is the average length of the words across both startWords and targetWords, N is the number of startWords, and M is the number of targetWords. This complexity arises because the code iterates over each character of every word in startWords to create bit masks (L * N operations), and then for each word in targetWords, it does the same to get a bitmask and tries to find if there is any word in startWords that can be turned into this target word by adding one letter (L * M operations).

The space complexity is O(N) because we store bit masks of all startWords in a set s. If k is the size of the alphabet and all words are of length L, the actual bit masks would take up L bits each, but since a set of integers is used here and the letters are mapped to bit indices, it only matters how many different bit masks there are, which corresponds to the number of startWords (or N).

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