Largest Merge Of Two Strings - Leetcode Solution

Problem Statement:

Given two strings word1 and word2, return the largest string such that every substring in the string can be rearranged to form one of the words in the input.

If there is more than one answer, return any of them. If there is no answer, return an empty string.

Example 1: Input: word1 = "abcabc", word2 = "abdcaba" Output: "abcabcabdcaba"

Example 2: Input: word1 = "ab", word2 = "cd" Output: "abcd"

Example 3: Input: word1 = "a", word2 = "a" Output: "a"

Example 4: Input: word1 = "a"*1000, word2 = "b"*1000 Output: ""

Approach:

The key concept used in this problem is to break the strings into smaller substrings and then combine them to obtain the largest possible string. In order to do this efficiently, we can use a graph representation where the nodes represent the substrings and the edges represent the possible ordering of these substrings.

We can find all the valid substrings of the given input strings word1 and word2 using a recursive function. For each substring, we can calculate its frequency count and store it along with the string in a dictionary.

Next, we can create a directed graph where each node represents a substring and there is an edge from node A to node B if and only if we can form the string B by merging the substrings represented by the nodes in the path from A to B. This can be done by comparing the frequency count of each character in the two substrings represented by the nodes.

Finally, we can traverse this graph to find the longest path from a starting node to an ending node. The longest path will represent the largest string that can be formed by merging the given input strings.

Solution:

def largestMerge(word1: str, word2: str) -> str:
    def generate_substrings(s: str, substrings: set):
        if s:
            substrings.add(s)
            generate_substrings(s[1:], substrings)
            generate_substrings(s[:-1], substrings)
    w1_substrings, w2_substrings = set(), set()
    generate_substrings(word1, w1_substrings)
    generate_substrings(word2, w2_substrings)
    substring_counts = {}
    for s in w1_substrings | w2_substrings:
        counts = [0] * 26
        for c in s:
            counts[ord(c) - ord('a')] += 1
        substring_counts[s] = tuple(counts)
    graph = {s: set() for s in substring_counts}
    for s1 in substring_counts:
        for s2 in substring_counts:
            if s1 != s2 and substring_counts[s1] == substring_counts[s2]:
                graph[s1].add(s2)
    longest_path = ''
    for node in graph:
        stack, visited = [(node, node)], set()
        while stack:
            curr, path = stack.pop()
            if curr in visited:
                continue
            visited.add(curr)
            if len(path) > len(longest_path) or (
               len(path) == len(longest_path) and path > longest_path):
                longest_path = path
            for neighbor in graph[curr]:
                stack.append((neighbor, path + neighbor))
    return longest_path

Explanation:

1. First, we define a wrapper function largestMerge that takes in two strings as input and returns the largest possible merged string.

2. We define a recursive helper function generate_substrings that takes in a string and a set of substrings. This function generates all possible substrings of the input string and adds them to the set of substrings.

3. We generate the set of substrings for both word1 and word2 using the generate_substrings function and store them in w1_substrings and w2_substrings, respectively.

4. Next, we generate a dictionary substring_counts that stores the frequency count of each character in each substring. For this, we loop through all the substrings in the union of w1_substrings and w2_substrings and calculate their frequency counts using a list of size 26 (one for each lowercase character), initialized with zeros. We then store the resulting tuple of frequency counts in the substring_counts dictionary.

5. We create a directed graph graph where each node represents a substring and there is an edge from node A to node B if and only if we can form the string B by merging the substrings represented by the nodes in the path from A to B. For this, we loop through all pairs of substrings and check if they have the same frequency counts, indicating that they can be combined to form a larger string. If so, we add an edge from the node representing the first substring to the node representing the second substring.

6. We initialize the variable longest_path to the empty string.

7. We loop through each node in the graph and use a depth-first traversal with a stack to find the longest path from that node to another node in the graph. For this, we start at the current node, push it onto the stack along with its string representation, and pop nodes and their paths from the stack until it is empty. We skip any nodes that have been already visited.

8. If the length of the current path is longer than that of longest_path, or if the lengths are the same but the current path is lexicographically larger, we update longest_path.

9. Finally, we return the longest_path.

Time Complexity:

The time complexity of the solution is O(N^2 log N) where N is the total number of substrings obtained from both the input strings. This is because we need to generate all possible substrings and then compare them with each other to create the graph. The DFS traversal takes O(N log N) time, as there can be at most N nodes and each node can have up to N neighbors in the worst case.

Space Complexity:

The space complexity of the solution is O(N^2) where N is the total number of substrings obtained from both the input strings. This is because we need to store all possible substrings and their frequency counts, as well as the graph of all possible paths. However, this can be optimized by using a hashmap instead of a set to store the substrings in the graph.

1