#2800

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Shortest String That Contains Three Strings

Find the shortest string containing three given strings using a greedy approach while ensuring it is lexicographically smallest.

string greedy enumeration

Problem Statement

Given three strings a, b, and c, you need to find the shortest string that contains all three as substrings. The solution must prioritize efficiency and return the lexicographically smallest string when there are multiple valid results.

You are allowed to overlap the strings as much as possible to minimize the length of the resulting string. The final solution should satisfy the constraints of containing all three strings while being the smallest possible in lexicographical order.

Examples

Example 1

Input: a = "abc", b = "bca", c = "aaa"

Output: "aaabca"

We show that "aaabca" contains all the given strings: a = ans[2...4], b = ans[3..5], c = ans[0..2]. It can be shown that the length of the resulting string would be at least 6 and "aaabca" is the lexicographically smallest one.

Example 2

Input: a = "ab", b = "ba", c = "aba"

Output: "aba"

We show that the string "aba" contains all the given strings: a = ans[0..1], b = ans[1..2], c = ans[0..2]. Since the length of c is 3, the length of the resulting string would be at least 3. It can be shown that "aba" is the lexicographically smallest one.

Constraints

1 <= a.length, b.length, c.length <= 100
a, b, c consist only of lowercase English letters.

Solution Approach

Greedy String Combination

The greedy approach involves combining strings by overlapping as much as possible. For each pair of strings, try to find the maximum overlap where one string's suffix matches another's prefix. This reduces the length of the resultant string.

Enumeration of Possible Combinations

Enumerate all possible orders of the three strings to ensure that all combinations are checked. Each order is processed to find the shortest valid string. The lexicographically smallest result can be identified after testing all orders.

Invariant Validation

At every step of the greedy combination, validate the invariant that all three strings are present as substrings. If any of them is not included, the combination is invalid, and the process needs to continue with the next combination or overlap.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the approach used for finding overlaps and enumerating combinations. Space complexity varies with the number of string combinations stored and processed during the calculation of overlaps.

What Interviewers Usually Probe

The candidate demonstrates an understanding of greedy string matching and overlap techniques.
The candidate efficiently handles different combinations of string overlaps to minimize the final string length.
The candidate ensures that the lexicographical order is maintained in case of multiple valid results.

Common Pitfalls or Variants

Common pitfalls

Failing to check all possible combinations of string orders, leading to missing the lexicographically smallest result.
Not handling string overlaps correctly, resulting in unnecessarily long strings.
Not validating the invariant at each step, potentially leading to invalid results.

Follow-up variants

Modify the problem to work with more than three strings and find the shortest string containing all of them.
Implement a solution where string overlaps are not allowed and each string must be added fully to the result.
Change the problem to return the longest valid string instead of the shortest one while still containing all substrings.

FAQ

What is the pattern used in the shortest string containing three strings problem?

The problem follows a greedy choice pattern where string overlaps are maximized while maintaining an invariant that all strings are present as substrings.

How do I ensure the shortest string is also lexicographically smallest?

You need to check all possible combinations of string orders and validate that the resultant string is the shortest and lexicographically smallest.

Can I solve the problem without using enumeration of string combinations?

No, enumeration of all combinations is necessary to ensure the shortest valid string is found, especially when considering lexicographical order.

What is a common mistake when implementing the greedy approach for this problem?

A common mistake is failing to correctly overlap the strings, which results in unnecessarily long strings that do not satisfy the problem requirements.

What does invariant validation mean in this context?

Invariant validation ensures that all three input strings are present as substrings in the final string. If any string is missing, the current combination is invalid.

terminal

Solution

Solution 1: Enumeration

We enumerate all permutations of the three strings, and for each permutation, we merge the three strings to find the shortest string with the smallest lexicographical order.

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class Solution:
    def minimumString(self, a: str, b: str, c: str) -> str:
        def f(s: str, t: str) -> str:
            if s in t:
                return t
            if t in s:
                return s
            m, n = len(s), len(t)
            for i in range(min(m, n), 0, -1):
                if s[-i:] == t[:i]:
                    return s + t[i:]
            return s + t

        ans = ""
        for a, b, c in permutations((a, b, c)):
            s = f(f(a, b), c)
            if ans == "" or len(s) < len(ans) or (len(s) == len(ans) and s < ans):
                ans = s
        return ans

Solution 2: Enumeration + KMP

We can use the KMP algorithm to optimize the string merging process.

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class Solution:
    def minimumString(self, a: str, b: str, c: str) -> str:
        def f(s: str, t: str) -> str:
            if s in t:
                return t
            if t in s:
                return s
            m, n = len(s), len(t)
            for i in range(min(m, n), 0, -1):
                if s[-i:] == t[:i]:
                    return s + t[i:]
            return s + t

        ans = ""
        for a, b, c in permutations((a, b, c)):
            s = f(f(a, b), c)
            if ans == "" or len(s) < len(ans) or (len(s) == len(ans) and s < ans):
                ans = s
        return ans

Continue Practicing

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