What is the time complexity of Find the Shortest Superstring?

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

What is the space complexity of Find the Shortest Superstring?

The space complexity is O(n * 2^n).

What data structures are used to solve Find the Shortest Superstring?

This problem uses Array, String, Dynamic Programming, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find the Shortest Superstring?

Insight 1: Dynamic programming with bitmasking is suitable because the number of words is small (<= 12), allowing us to represent the set of used words with a bitmask. Insight 2: Calculating the overlap between each pair of words beforehand avoids redundant computations within the dynamic programming loop.

Which companies ask LeetCode 943. Find the Shortest Superstring in interviews?

This problem is frequently asked at DE Shaw.

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Find the Shortest Superstring - LeetCode 943 Solution

Q: How to solve Find the Shortest Superstring?

The solution uses dynamic programming with bitmasking to find the optimal order of strings. It first computes the maximum overlap between each pair of strings `words[i]` and `words[j]`. Then, it uses a DP table `dp[mask][i]` to store the maximum overlap achieved by including the strings represented by the bitmask `mask` and ending with the string `words[i]`. The `parent[mask][i]` table stores the previous string in the optimal path. Finally, it reconstructs the optimal path and concatenates the strings, considering overlaps to create the shortest superstring.

Q: Which companies ask LeetCode 943. Find the Shortest Superstring in interviews?

This problem is frequently asked at DE Shaw.

Find the Shortest Superstring - Complete Solution Guide

Find the Shortest Superstring is LeetCode problem 943, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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","

Detailed Explanation

The problem requires finding the shortest possible string (superstring) that contains all the strings in a given array of strings as substrings. The input is an array of strings called `words`. The output is the shortest superstring containing all input strings. We are guaranteed that no string in `words` is a substring of another string in `words`. If there are multiple shortest superstrings, any one of them can be returned. The number of words is limited to 12, and each word is at most 20 characters long.

Solution Approach

The solution uses dynamic programming with bitmasking to find the optimal order of strings. It first computes the maximum overlap between each pair of strings `words[i]` and `words[j]`. Then, it uses a DP table `dp[mask][i]` to store the maximum overlap achieved by including the strings represented by the bitmask `mask` and ending with the string `words[i]`. The `parent[mask][i]` table stores the previous string in the optimal path. Finally, it reconstructs the optimal path and concatenates the strings, considering overlaps to create the shortest superstring.

Step-by-Step Algorithm

Step 1: Calculate the overlaps between all pairs of words. The `overlaps[i][j]` stores the length of the longest suffix of `words[i]` that is also a prefix of `words[j]`.
Step 2: Initialize the DP table `dp` and `parent` table. `dp[mask][i]` stores the maximum overlap obtained using the strings represented by the mask `mask` and ending with word `i`. `parent[mask][i]` stores the index of the word preceding word `i` in the shortest superstring formed by words in mask `mask` ending with word `i`.
Step 3: Iterate through all possible masks from 1 to 2^n - 1. For each mask, iterate through all words `i` included in the mask. Then, iterate through all words `j` in the previous mask (`mask` XOR `1 << i`). Update `dp[mask][i]` and `parent[mask][i]` if a better overlap is found by appending word `i` after word `j`.
Step 4: Find the maximum overlap in the final mask `(1 << n) - 1`. This gives the maximum overlap across all permutations.
Step 5: Backtrack through the `parent` table to reconstruct the optimal order of strings in the superstring.
Step 6: Construct the shortest superstring by concatenating the words in the optimal order, considering the overlaps between adjacent words.

Key Insights

Insight 1: Dynamic programming with bitmasking is suitable because the number of words is small (<= 12), allowing us to represent the set of used words with a bitmask.
Insight 2: Calculating the overlap between each pair of words beforehand avoids redundant computations within the dynamic programming loop.
Insight 3: The problem can be viewed as finding the optimal order of the strings in `words` to minimize the length of their concatenation, considering overlaps.

Complexity Analysis

Time Complexity: O(n^2 * 2^n)

Space Complexity: O(n * 2^n)

Topics

This problem involves: Array, String, Dynamic Programming, Bit Manipulation, Bitmask.

Companies

Asked at: DE Shaw.

Find the Shortest Superstring | LeetCode 943

Hard

Array String Dynamic Programming Bit Manipulation Bitmask

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public String shortestSuperstring(String[] words) {
        int n = words.length;

        int[][] overlaps = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    continue;
                }
                for (int k = Math.min(words[i].length(), words[j].length()); k > 0; k--) {
                    if (words[i].endsWith(words[j].substring(0, k))) {
                        overlaps[i][j] = k;
                        break;
                    }
                }
            }
        }

        int[][] dp = new int[1 << n][n];
        int[][] parent = new int[1 << n][n];
        for (int i = 0; i < (1 << n); i++) {
            for (int j = 0; j < n; j++) {
                parent[i][j] = -1;
            }
        }

        for (int mask = 1; mask < (1 << n); mask++) {
            for (int i = 0; i < n; i++) {
                if ((mask >> i) % 2 == 1) {
                    int prevMask = mask ^ (1 << i);
                    if (prevMask == 0) {
                        continue;
                    }

                    for (int j = 0; j < n; j++) {
                        if ((prevMask >> j) % 2 == 1) {
                            int val = dp[prevMask][j] + overlaps[j][i];
                            if (val > dp[mask][i]) {
                                dp[mask][i] = val;
                                parent[mask][i] = j;
                            }
                        }
                    }
                }
            }
        }

        int maxOverlap = -1;
        int lastIdx = -1;
        int finalMask = (1 << n) - 1;

        for (int i = 0; i < n; i++) {
            if (dp[finalMask][i] > maxOverlap) {
                maxOverlap = dp[finalMask][i];
                lastIdx = i;
            }
        }

        List<Integer> path = new ArrayList<>();
        int currMask = finalMask;
        int currIdx = lastIdx;

        while (currIdx != -1) {
            path.add(currIdx);
            int prevIdx = parent[currMask][currIdx];
            currMask ^= (1 << currIdx);
            currIdx = prevIdx;
        }

        List<String> res = new ArrayList<>();
        for (int i = path.size() - 1; i >= 0; i--) {
            res.add(words[path.get(i)]);
        }

        StringBuilder sb = new StringBuilder(res.get(0));
        for (int i = 1; i < n; i++) {
            int prev = path.get(path.size() - 1 - (i - 1));
            int curr = path.get(path.size() - 1 - i);
            int overlapLen = overlaps[prev][curr];
            sb.append(words[curr].substring(overlapLen));
        }

        return sb.toString();
    }
}

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Problem Info

Description

Problem 943: Find the Shortest Superstring - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Bit Manipulation Bitmask

Companies

DE Shaw

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:943

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: Calculate the overlaps between all pairs of words. The `overlaps[i][j]` stores the length of the longest suffix of `words[i]` that is also a prefix of `words[j]`.
2
Step 2: Initialize the DP table `dp` and `parent` table. `dp[mask][i]` stores the maximum overlap obtained using the strings represented by the mask `mask` and ending with word `i`. `parent[mask][i]` stores the index of the word preceding word `i` in the shortest superstring formed by words in mask `mask` ending with word `i`.
3
Step 3: Iterate through all possible masks from 1 to 2^n - 1. For each mask, iterate through all words `i` included in the mask. Then, iterate through all words `j` in the previous mask (`mask` XOR `1 << i`). Update `dp[mask][i]` and `parent[mask][i]` if a better overlap is found by appending word `i` after word `j`.
4
Step 4: Find the maximum overlap in the final mask `(1 << n) - 1`. This gives the maximum overlap across all permutations.
5
Step 5: Backtrack through the `parent` table to reconstruct the optimal order of strings in the superstring.
6
Step 6: Construct the shortest superstring by concatenating the words in the optimal order, considering the overlaps between adjacent words.

💡

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 * 2^n)

💾

Space Complexity

O(n * 2^n)

📝

Why?

The time complexity is dominated by the dynamic programming step. The outer loop iterates through all possible masks (2^n), and the inner loops iterate through all possible pairs of words (n^2). Computing overlaps takes O(n^2 * L) in the worst case (where L is the maximum length of a word), but it is done only once as preprocessing. The backtracking takes O(n). Thus, overall Time complexity is O(n^2 * 2^n). The space complexity is determined by the `dp` and `parent` tables, which are of size O(n * 2^n).

📖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

Greedy Approach

A greedy approach could involve repeatedly merging the two strings with the maximum overlap until only one string remains. However, this approach doesn't guarantee the shortest superstring, as local optimal choices might not lead to the global optimum.

Time: O(n^2 * L * n), where n is number of words and L is average length of wordsSpace: O(L * n)

Trade-offs: Faster than DP approach, but may not always give the shortest superstring. Useful if approximate solution is sufficient.

Brute-force

Generate all possible permutations of the input `words` and check if any permutation results in smallest length superstring. This is very inefficient.

Time: O(n! * n * L), where n is number of words, and L is average word length. n! for generating permutations and O(n*L) for constructing the string.Space: O(L * n)

Trade-offs: Simple to implement but highly inefficient and unusable for even small inputs, due to factorial time complexity.

✨

Key Insights

1Insight 1: Dynamic programming with bitmasking is suitable because the number of words is small (<= 12), allowing us to represent the set of used words with a bitmask.
2Insight 2: Calculating the overlap between each pair of words beforehand avoids redundant computations within the dynamic programming loop.
3Insight 3: The problem can be viewed as finding the optimal order of the strings in `words` to minimize the length of their concatenation, considering overlaps.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: Empty input array. The problem statement doesn't explicitly mention this, but the code should handle the case gracefully (which it does by not triggering loops).
Edge case 2: Input array with a single word. The code correctly returns the single word as the shortest superstring.
Edge case 3: Very long words compared to number of words. The overlap computation cost increases, but within problem constraint limits, the bitmask DP dominates.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not pre-computing the overlaps efficiently. This leads to redundant calculations within the DP loop, increasing the time complexity.

✗Common mistake 2: Incorrectly reconstructing the path from the `parent` table. This can lead to an incorrect superstring.

✗Common mistake 3: Failing to handle edge cases like an empty input array or an array with a single word.

✅

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: Genome sequencing. Finding the shortest sequence that contains multiple DNA fragments as substrings is crucial in bioinformatics.

🏦Real-world scenario 2: Data compression. Finding overlaps between data segments and merging them can help in reducing the overall size of the data.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: July 14, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find the Shortest Superstring"?

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

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