# Solving Leetcode Interviews in Seconds with AI: Find the Shortest Superstring


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "943" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an array of strings words, return the smallest string that contains each string in words as a substring. If there are multiple valid strings of the smallest length, return any of them. You may assume that no string in words is a substring of another string in words.   Example 1:  Input: words = ["alex","loves","leetcode"] Output: "alexlovesleetcode" Explanation: All permutations of "alex","loves","leetcode" would also be accepted.  Example 2:  Input: words = ["catg","ctaagt","gcta","ttca","atgcatc"] Output: "gctaagttcatgcatc"    Constraints:  1 <= words.length <= 12 1 <= words[i].length <= 20 words[i] consists of lowercase English letters. All the strings of words are unique.  

	# Explanation
	Here's a breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Calculate the overlap (prefix/suffix matching) between all pairs of words. This overlap information is stored in a matrix.
    *   Use dynamic programming (with bitmasking) to find the shortest string that includes all words. The bitmask represents which words have been included so far.
    *   Reconstruct the shortest string by backtracking through the DP table.

*   **Complexity:**
    *   Runtime: O(n<sup>2</sup> * 2<sup>n</sup>), where n is the number of words.
    *   Storage: O(n * 2<sup>n</sup>)

	
	# Code
	```python
	def shortest_superstring(words):
    n = len(words)

    # Calculate overlaps
    overlaps = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(n):
            if i != j:
                for k in range(min(len(words[i]), len(words[j])), -1, -1):
                    if words[i].endswith(words[j][:k]):
                        overlaps[i][j] = k
                        break

    # DP: dp[mask][i] is the shortest string covering words represented by mask, ending with word i
    dp = [[""] * n for _ in range(1 << n)]
    parent = [[0] * n for _ in range(1 << n)]  # Store parent word for backtracking

    # Initialize base cases: single words
    for i in range(n):
        dp[1 << i][i] = words[i]

    # Iterate through all possible subsets of words
    for mask in range(1, 1 << n):
        for i in range(n):
            if (mask & (1 << i)) != 0:  # If word i is in the subset
                for j in range(n):
                    if i != j and (mask & (1 << j)) != 0:  # If word j is also in the subset and not the same as i
                        new_string = dp[mask ^ (1 << i)][j] + words[i][overlaps[j][i]:]
                        if dp[mask][i] == "" or len(new_string) < len(dp[mask][i]):
                            dp[mask][i] = new_string
                            parent[mask][i] = j

    # Find the shortest string among all possible ending words
    min_len = float('inf')
    last_word = -1
    for i in range(n):
        if len(dp[(1 << n) - 1][i]) > 0 and len(dp[(1 << n) - 1][i]) < min_len:
            min_len = len(dp[(1 << n) - 1][i])
            last_word = i

    # Reconstruct the shortest superstring by backtracking
    result = ""
    mask = (1 << n) - 1
    while mask > 0:
        result = dp[mask][last_word]
        temp = last_word
        last_word = parent[mask][last_word]
        mask ^= (1 << temp)

    return result
	```