What is the time complexity of Shortest String That Contains Three Strings?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Shortest String That Contains Three Strings?

This problem uses String, Greedy, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shortest String That Contains Three Strings?

Insight 1: The order in which the strings are concatenated/merged matters significantly. To find the absolute shortest string, all possible permutations of the input strings must be considered. Insight 2: Overlapping substrings can reduce the length of the final string. The algorithm needs to efficiently identify and utilize the maximum possible overlap between strings during merging.

Which companies ask LeetCode 2800. Shortest String That Contains Three Strings in interviews?

This problem is frequently asked at DE Shaw.

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Shortest String That Contains Three Strings - LeetCode 2800 Solution

Shortest String That Contains Three Strings - Complete Solution Guide

Shortest String That Contains Three Strings is LeetCode problem 2800, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given three strings a , b , and c , your task is to find a string that has the minimum length and contains all three strings as substrings . If there are multiple such strings, return the lexicographically smallest one. Return a string denoting the answer to the problem. Notes A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b . A

Detailed Explanation

The problem requires finding the shortest string that contains three given strings (a, b, and c) as substrings. If multiple shortest strings exist, the lexicographically smallest one should be returned. The input strings consist of lowercase English letters and have lengths between 1 and 100, inclusive.

Solution Approach

The solution uses a brute-force approach combined with a merging function. It generates all possible permutations of the three input strings. For each permutation, it merges the strings sequentially, maximizing the overlap between each pair. Finally, it selects the shortest resulting string among all permutations, and if there are multiple shortest strings, it chooses the lexicographically smallest one.

Step-by-Step Algorithm

Step 1: Define a `merge` function that takes two strings (s1, s2) as input and returns their merged string. This function finds the maximum overlap between the suffix of s1 and the prefix of s2. If s2 is a substring of s1, return s1. Otherwise, identify the largest k (from min(len(s1), len(s2)) down to 0) such that the last k characters of s1 match the first k characters of s2. The merged string is then s1 + the remaining part of s2 (s2[k:]).
Step 2: Generate all possible permutations of the three input strings (a, b, c). In Python, `itertools.permutations` is used. In Java and C++, the permutations are generated manually. In C, this is done by directly listing out each permutation.
Step 3: For each permutation, apply the `merge` function twice to combine the strings. For example, for permutation (s1, s2, s3), the merging occurs as `merge(merge(s1, s2), s3)`.
Step 4: Store all the resulting merged strings (candidates).
Step 5: Iterate through the candidate strings and find the one with the shortest length. If multiple strings have the same shortest length, choose the lexicographically smallest one using a string comparison function (e.g., `<` operator in C++, `compareTo` in Java, `strcmp` in C).
Step 6: Return the selected string.

Key Insights

Insight 1: The order in which the strings are concatenated/merged matters significantly. To find the absolute shortest string, all possible permutations of the input strings must be considered.
Insight 2: Overlapping substrings can reduce the length of the final string. The algorithm needs to efficiently identify and utilize the maximum possible overlap between strings during merging.
Insight 3: When strings have the same length, the lexicographically smallest string needs to be selected. This requires using a string comparison function.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy, Enumeration.

Companies

Asked at: DE Shaw.

Shortest String That Contains Three Strings | LeetCode 2800

Medium

String Greedy Enumeration

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public String minimumString(String a, String b, String c) {
        
        // Helper function to merge two strings
        class Helper {
            String merge(String s1, String s2) {
                if (s1.contains(s2)) {
                    return s1;
                }

                for (int k = Math.min(s1.length(), s2.length()); k >= 0; k--) {
                    if (s1.endsWith(s2.substring(0, k))) {
                        return s1 + s2.substring(k);
                    }
                }

                return s1 + s2;
            }
        }
        
        Helper helper = new Helper();

        String[] strings = {a, b, c};
        String[] candidates = new String[6];
        int candidateIndex = 0;

        // Generate all permutations
        String[] p = new String[3];
        
        p[0] = strings[0];
        p[1] = strings[1];
        p[2] = strings[2];
        
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);

        p[0] = strings[0];
        p[1] = strings[2];
        p[2] = strings[1];
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);
        
        p[0] = strings[1];
        p[1] = strings[0];
        p[2] = strings[2];
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);
        
        p[0] = strings[1];
        p[1] = strings[2];
        p[2] = strings[0];
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);
        
        p[0] = strings[2];
        p[1] = strings[0];
        p[2] = strings[1];
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);
        
        p[0] = strings[2];
        p[1] = strings[1];
        p[2] = strings[0];
        candidates[candidateIndex++] = helper.merge(helper.merge(p[0], p[1]), p[2]);


        // Find the best candidate
        String bestCandidate = candidates[0];
        for (int i = 1; i < 6; i++) {
            if (candidates[i].length() < bestCandidate.length()) {
                bestCandidate = candidates[i];
            } else if (candidates[i].length() == bestCandidate.length() && candidates[i].compareTo(bestCandidate) < 0) {
                bestCandidate = candidates[i];
            }
        }

        return bestCandidate;
    }
}

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Problem Info

Description

Problem 2800: Shortest String That Contains Three Strings - Detailed explanation and solution approaches.

Topics

String Greedy Enumeration

Companies

DE Shaw

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2800

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Define a `merge` function that takes two strings (s1, s2) as input and returns their merged string. This function finds the maximum overlap between the suffix of s1 and the prefix of s2. If s2 is a substring of s1, return s1. Otherwise, identify the largest k (from min(len(s1), len(s2)) down to 0) such that the last k characters of s1 match the first k characters of s2. The merged string is then s1 + the remaining part of s2 (s2[k:]).
2
Step 2: Generate all possible permutations of the three input strings (a, b, c). In Python, `itertools.permutations` is used. In Java and C++, the permutations are generated manually. In C, this is done by directly listing out each permutation.
3
Step 3: For each permutation, apply the `merge` function twice to combine the strings. For example, for permutation (s1, s2, s3), the merging occurs as `merge(merge(s1, s2), s3)`.
4
Step 4: Store all the resulting merged strings (candidates).
5
Step 5: Iterate through the candidate strings and find the one with the shortest length. If multiple strings have the same shortest length, choose the lexicographically smallest one using a string comparison function (e.g., `<` operator in C++, `compareTo` in Java, `strcmp` in C).
6
Step 6: Return the selected string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `merge` function has a time complexity of O(n) because it iterates up to `n` times, where `n` is the length of the strings being merged, when searching for the optimal overlap. Because we have to merge, at most, 3 strings, the final length of the output string can have at most a length of 3n, thus justifying the O(n) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Aho-Corasick Algorithm

Use the Aho-Corasick algorithm to build a state machine representing the three input strings. Then, traverse the state machine to construct the shortest string containing all three. This approach could potentially optimize the substring search.

Time: O(N + M), where N is the total length of all three strings and M is the length of the shortest combined string.Space: O(N), where N is the total length of all three strings.

Trade-offs: More complex to implement than the brute-force approach, but may be more efficient for larger input strings due to optimized substring search. The constant factors might make it slower for these small strings.

✨

Key Insights

1Insight 1: The order in which the strings are concatenated/merged matters significantly. To find the absolute shortest string, all possible permutations of the input strings must be considered.
2Insight 2: Overlapping substrings can reduce the length of the final string. The algorithm needs to efficiently identify and utilize the maximum possible overlap between strings during merging.
3Insight 3: When strings have the same length, the lexicographically smallest string needs to be selected. This requires using a string comparison function.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: One string is a substring of another. The `merge` function handles this correctly by simply returning the longer string, as required by the problem.
Edge case 2: The strings have no overlap at all. The `merge` function will correctly append the strings without any overlap.
Edge case 3: Empty strings (although the constraints do not permit these). The `merge` function still handles the case of empty strings when merging prefixes and suffixes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all permutations of the input strings. This will likely result in a string that isn't the shortest or lexicographically smallest.

✗Common mistake 2: Incorrectly implementing the merge function, especially when determining the maximum overlap. Failing to handle the case where one string is a substring of another can also lead to errors.

✗Common mistake 3: Forgetting to handle the case of equal length strings when finding the lexicographically smallest result. Using the wrong comparison operation can also cause this.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Bioinformatics - finding the shortest DNA sequence that contains specific gene sequences.

🏦Real-world scenario 2: Data compression - identifying overlapping segments in data to create a more compact representation. This is a simplified example, as real compression algorithms are far more complex.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Shortest String That Contains Three Strings"?

For this medium-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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