Problem Description

You are given two strings, s1 and s2. Return the shortest possible string that contains both s1 and s2 as substrings. If there are multiple valid answers, return any one of them.

A substring is a contiguous sequence of characters within a string.

To solve this, construct the shortest string that contains both s1 and s2 by maximizing overlap. Try to overlap the suffix of one string with the prefix of the other. Consider these scenarios:

1. If s1 is already a substring of s2, then s2 itself is a valid answer. The same applies if s2 is a substring of s1.
2. Try to overlap the end of s1 with the beginning of s2 as much as possible and concatenate the rest.
3. Similarly, try to overlap the end of s2 with the beginning of s1.
4. If there is no overlap, simply concatenate s1 and s2.

The answer should be the shortest string containing both as substrings. If multiple shortest strings exist, you can return any of them.

Intuition

To build the shortest string that contains both s1 and s2 as substrings, try to merge them with as much overlap as possible. Since a substring is a continuous part of a string, look for places where the end of one string matches the beginning of the other. For example, if the suffix of s1 matches the prefix of s2, those characters don’t need to appear twice. The same idea works in reverse.

If neither string contains the other and there’s no overlap, just stick them together (i.e., s1 + s2). But if any overlap exists — like some ending of s1 is the same as the beginning of s2 — use that to make the merged string shorter. The closer they are “lined up,” the shorter your result will be. This way, the problem becomes searching for the maximum length of overlap between the two strings in both possible orders.

Solution Approach

To find the shortest string containing both s1 and s2 as substrings, follow these steps:

1. Check if either string contains the other:

   • If s1 is already a substring of s2, return s2.
   • If s2 is already a substring of s1, return s1.
   • This case gives the shortest possible answer immediately.

2. Find maximum suffix-prefix overlap:

   • Try to overlap the end of s1 with the start of s2 as much as possible.
   • For each suffix of s1, check if it matches the corresponding prefix of s2.
   • If s2.startswith(s1[i:]) is True, then the overlap is s1[i:], and you can combine them as s1[:i] + s2.
   • This reduces the total length by the length of the overlap.

3. Find maximum prefix-suffix overlap (reverse order):

   • Also check if the beginning of s1 matches the end of s2.
   • For each prefix of s1, check if it is equal to the corresponding suffix of s2.
   • If s2.endswith(s1[:m - i]) is True, then overlap is s1[:m - i], and you can combine as s2 + s1[m - i:].

4. No overlap:

   • If neither of the above approaches yielded an overlap, just return the two strings concatenated: s1 + s2.

This process ensures you always get the shortest possible result because it checks all ways the strings might overlap, and stops at the largest overlap.

The major pattern here is checking overlaps using string matching, by sliding through potential suffixes and prefixes of both strings. This leads to a time complexity of O(m * n), where m and n are the lengths of the two strings.

Here’s the basic logic shown in Python-style pseudocode:

def shortestSuperstring(s1, s2):
    if s1 in s2:
        return s2
    if s2 in s1:
        return s1

    # Check for suffix-prefix overlap
    for i in range(len(s1)):
        if s2.startswith(s1[i:]):
            return s1[:i] + s2

    # Check for prefix-suffix overlap
    for i in range(len(s1)):
        if s2.endswith(s1[:len(s1)-i]):
            return s2 + s1[len(s1)-i:]

    # No overlap
    return s1 + s2

This approach is efficient and directly matches the described logic in the solution code.

Example Walkthrough

Let's walk through the approach using an example.

Suppose:

• s1 = "abcde"
• s2 = "cdefg"

Step 1: Check for direct containment

• "abcde" is not a substring of "cdefg"
• "cdefg" is not a substring of "abcde"
• So, move to the next step.

Step 2: Find maximum suffix-prefix overlap (s1 suffix matches s2 prefix)

Check overlaps by sliding through the suffixes of s1:

• i = 0: Does "cdefg" start with "abcde"? No
• i = 1: Does "cdefg" start with "bcde"? No
• i = 2: Does "cdefg" start with "cde"? Yes! Overlap is "cde" (length 3)

So we can merge by taking the non-overlapping prefix of s1 and the entirety of s2:

• Result: "ab" (from s1[:2]) + "cdefg" = "abcdefg"

Step 3: Check maximum prefix-suffix overlap (reverse order)

We already found an overlap, but the approach would also check the other way:

Check if "abcde" starts with any suffix of "cdefg":

• i = 0: Does "abcde" start with "cdefg"? No
• i = 1: Does "abcde" start with "defg"? No
• i = 2: Does "abcde" start with "efg"? No
• ... No better overlap found in this direction.

Step 4: If no overlap was found, concatenate: Not needed here; we've already found the shortest possible superstring.

Summary: The merged shortest string is "abcdefg".

This walk-through demonstrates checking for substring containment, sliding suffix-prefix overlaps, and reversing the direction—all to produce the minimal superstring.

Solution Implementation

Time and Space Complexity

• Time Complexity: The most significant loops in the code are the two for loops which each iterate at most m times (m = len(s1)). Inside each loop, string slicing and the startswith or endswith operations are called, both of which take up to O(n) time (n = max(len(s1), len(s2))). Thus, the time complexity per loop is O(m * n). Since m <= n, this is O(n^2) overall.

• Space Complexity: The space required is mainly for temporary substrings created during slicing and concatenation, each up to length n, resulting in a space complexity of O(n).