Problem Description

The problem presents a task where we need to merge two given strings word1 and word2 into one new string merge. The goal is to create the lexicographically largest string possible. The process of merging is defined by repeatedly taking the first character from either word1 or word2 and appending it to merge, then removing that character from the string it was taken from. The lexicographically largest string means that if you sort all possible merge strings, the one we want would appear last. It should be constructed in such a way that at every choice, if possible, the character that will make merge lexicographically larger should be chosen.

We're asked to implement a function that, given two strings word1 and word2, returns the lexicographically largest merge string that can be constructed from them.

Intuition

The intuition behind the solution is to always pick the lexicographically larger character to append to the merge string. However, simply comparing the characters at the current positions in word1 and word2 is not enough. We should look ahead because picking a character from one string might lead to a suboptimal result if the subsequent characters in the other string would create a lexicographically larger string.

The solution approach is to compare the substrings starting from the current characters of word1 and word2, not just the characters themselves. This comparison tells us which string leads to a lexicographically larger outcome if we were to take all remaining characters from it. Whenever the substring of word1 from the current index i is greater than the substring of word2 from the current index j, we append the character from word1 to merge, and vice versa.

We use a while loop to conduct this process repeatedly until one of the strings is empty. Once one of the strings is empty, there are no more decisions to be made—we simply append the remaining characters of the non-empty string to merge. The Python > operator is used for the comparison, which conveniently compares strings lexicographically. The .join() method is then used to combine the list of characters into a single string before returning it as the solution.

Learn more about Greedy and Two Pointers patterns.

Solution Approach

The implemented solution uses two pointers, i and j, which start at 0 corresponding to the first characters in word1 and word2 respectively. An empty list named ans is initialized to collect the characters that will form the merge string.

The main algorithm is composed of a while loop, which runs as long as there are characters left in both word1 and word2. Within this loop, the key operation is to compare the substrings of word1 starting from i and word2 starting from j. This is done with the expression word1[i:] > word2[j:].

• If word1[i:] is lexicographically larger than word2[j:], the first character of word1 at index i is appended to the ans list using ans.append(word1[i]), and the pointer i is incremented by 1 with i += 1.
• If word2[j:] is lexicographically larger or equal to word1[i:], the first character of word2 at index j is appended to the ans list using ans.append(word2[j]), and the pointer j is incremented by 1 with j += 1.

Once the while loop exits (meaning at least one of the strings is exhausted), the remaining characters from both strings (if any) are appended to the ans list using ans.append(word1[i:]) and ans.append(word2[j:]). These operations effectively concatenate the leftover substring to the merge string.

Finally, the merge string is constructed by joining the characters in the ans list with the "".join(ans) expression, which combines all elements of the list into a single string. The resulting string is then returned as the largest lexicographical merge that can be constructed from word1 and word2.

This solution makes use of simple data structures (strings and lists) and an algorithm that optimally decides which character to append to merge at each step, ensuring the lexicographically largest result.

Example Walkthrough

Let's assume we are given the following input strings:

word1 = "ace"
word2 = "bdf"

Our goal is to merge word1 and word2 into the lexicographically largest string possible as per the solution approach described. Here is a step-by-step walkthrough of how the algorithm will work with these inputs:

1. Initialize pointers i and j both to 0 and an empty list ans to collect characters.
2. Compare word1[0:] ("ace") with word2[0:] ("bdf").
3. Since 'ace' < 'bdf' lexicographically, we append the first character of word2 to ans (['b']), and increment j to 1.
4. Now compare word1[0:] ("ace") with word2[1:] ("df").
5. 'ace' < 'df' lexicographically, so append the first character of word2 at index j to ans (['b', 'd']), and increment j to 2.
6. Now compare word1[0:] ("ace") with word2[2:] ("f").
7. 'ace' > 'f' lexicographically, so append the first character of word1 ('a') to ans (['b', 'd', 'a']), and increment i to 1.
8. Compare word1[1:] ("ce") with word2[2:] ("f").
9. 'ce' > 'f' lexicographically, append the next character of word1 at index i to ans (['b', 'd', 'a', 'c']), and increment i to 2.
10. Compare word1[2:] ("e") with word2[2:] ("f").
11. 'e' < 'f' lexicographically, append the character of word2 at index j to ans (['b', 'd', 'a', 'c', 'f']), and increment j to 3.
12. word2 is now empty, so we append the remaining characters of word1 to ans.
13. Adding word1[2:] ("e") to ans gives us ['b', 'd', 'a', 'c', 'f', 'e'].
14. Join the characters in ans with "".join(ans) to get the final merged string.

The resulting merge string is "bdacfe", which is the lexicographically largest string constructible from the input word1 and word2.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be analyzed by looking at the operations inside the while loop and the operations that happen after the while loop.

1. The while loop runs until i < len(word1) and j < len(word2). At each iteration, it checks the lexicographical order of the suffixes starting at the current indices i in word1 and j in word2. Comparing the suffixes (word1[i:] > word2[j:]) is an O(m) operation, where m is the length of the longer suffix at each step because in the worst case, comparison could go on till the end of the string.

2. The loop runs up to len(word1) + len(word2) times in total since at each iteration at least one character is appended to ans.

Therefore, the worst-case time complexity is O((len(word1) + len(word2)) * m), where m is the length of the longer suffix at each step.

However, if we consider that string comparison in Python is done lexicographically and character by character, m will be the smaller of the two suffix lengths at each comparison point.

So a tighter bound considering the average lengths as average sizes of the compared suffixes, the time complexity would be O(n * k), where n is len(word1) + len(word2) and k is the average size of these suffixes during comparison operations.

Space Complexity

The space complexity can be analyzed by considering the extra space used by the algorithm.

1. The list ans can grow up to len(word1) + len(word2) characters in size.

2. The string slices word1[i:] and word2[j:], if implemented naively, could potentially create new strings each iteration, but in Python, slicing strings doesn't create copies, but rather, new references to the existing string's elements. So, these operations are O(1) in space.

Therefore, the space complexity is O(n), where n is len(word1) + len(word2) for the result list that is generated.

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