What is the time complexity of Shortest Uncommon Substring in an Array?

The optimal solution has a time complexity of O(N * L^3).

What is the space complexity of Shortest Uncommon Substring in an Array?

The space complexity is O(N * L^2).

What data structures are used to solve Shortest Uncommon Substring in an Array?

This problem uses Array, Hash Table, String, Trie. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shortest Uncommon Substring in an Array?

Insight 1: The core idea is to efficiently identify all substrings of each string and track their occurrences across all strings in the input array. Insight 2: A hash table (map or dictionary) is ideal for storing substrings and the indices of the strings they appear in. This allows for quick lookup of substring occurrences.

Which companies ask LeetCode 3076. Shortest Uncommon Substring in an Array in interviews?

This problem is frequently asked at Affirm, Airbnb, Moveworks.

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Shortest Uncommon Substring in an Array - LeetCode 3076 Solution

Shortest Uncommon Substring in an Array - Complete Solution Guide

Shortest Uncommon Substring in an Array is LeetCode problem 3076, 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

You are given an array arr of size n consisting of non-empty strings. Find a string array answer of size n such that: answer[i] is the shortest substring of arr[i] that does not occur as a substring in any other string in arr . If multiple such substrings exist, answer[i] should be the lexicographically smallest . And if no such substring exists, answer[i] should be an empty string. Return the array answer . Example 1: Input: arr = ["cab","ad","bad","c"] Output: ["ab","","ba",""] Explanation: We

Detailed Explanation

The problem requires finding, for each string in an input array `arr`, the shortest substring that is unique to that string, meaning it doesn't appear as a substring in any other string in `arr`. If multiple such substrings of the same length exist, the lexicographically smallest one should be selected. If no such substring exists, the result for that string should be an empty string. The output should be an array of strings, where each element corresponds to the solution for the corresponding string in the input array.

Solution Approach

The provided solution constructs a map (or dictionary) where keys are substrings and values are sets of indices of strings in which those substrings appear. Then, for each string in the input array, it iterates through its substrings, checking if the substring's index set in the map contains only the index of the current string. If so, it means the substring is unique to that string. If multiple unique substrings of the same length are found, the lexicographically smallest one is chosen. The algorithm finds the shortest unique substring for each string in the array and returns the array of these substrings.

Step-by-Step Algorithm

Step 1: Initialize a map (substring_indices) to store substrings as keys and sets of string indices as values. This map will track in which strings each substring appears.
Step 2: Iterate through each string `s` in the input array `arr` at index `i`.
Step 3: Generate all substrings of `s` with lengths ranging from 1 to the length of `s`.
Step 4: For each substring, add the index `i` to the set of indices associated with that substring in the `substring_indices` map.
Step 5: After processing all strings and their substrings, initialize an empty array `answer` to store the results.
Step 6: Iterate through each string `s` in the input array `arr` again, this time to find the shortest unique substring.
Step 7: For each string `s`, iterate through its substrings (again, in increasing order of length).
Step 8: Check if the current substring appears in the `substring_indices` map. If it does, check the size of the associated index set. If the size of the index set is 1 (meaning the substring appears only in the current string), add the substring to a list of `candidates`.
Step 9: If any `candidates` are found, select the lexicographically smallest substring and add it to the `answer` array. Break the inner loop to ensure that the shortest substring is chosen.
Step 10: If no unique substring is found for a string, add an empty string to the `answer` array.
Step 11: Return the `answer` array.

Key Insights

Insight 1: The core idea is to efficiently identify all substrings of each string and track their occurrences across all strings in the input array.
Insight 2: A hash table (map or dictionary) is ideal for storing substrings and the indices of the strings they appear in. This allows for quick lookup of substring occurrences.
Insight 3: Iterating through substrings in increasing order of length allows for early termination when a unique substring is found, ensuring we find the *shortest* unique substring first. Lexicographical ordering can be done by simply using min() on a list of candidate strings of equal length.
Insight 4: Caching already seen substrings of a string during substring generation avoids redundant calculations and insertions into the hash table, improving efficiency.

Complexity Analysis

Time Complexity: O(N * L^3)

Space Complexity: O(N * L^2)

Topics

This problem involves: Array, Hash Table, String, Trie.

Companies

Asked at: Affirm, Airbnb, Moveworks.

Shortest Uncommon Substring in an Array | LeetCode 3076

Medium

Array Hash Table String Trie

Time: O(N * L^3)Space: O(N * L^2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public String[] shortestSubstrings(String[] arr) {
        int n = arr.length;
        Map<String, Set<Integer>> substringIndices = new HashMap<>();

        for (int i = 0; i < n; i++) {
            String s = arr[i];
            Set<String> seenSubsInS = new HashSet<>();
            for (int length = 1; length <= s.length(); length++) {
                for (int k = 0; k <= s.length() - length; k++) {
                    String sub = s.substring(k, k + length);
                    if (!seenSubsInS.contains(sub)) {
                        if (!substringIndices.containsKey(sub)) {
                            substringIndices.put(sub, new HashSet<>());
                        }
                        substringIndices.get(sub).add(i);
                        seenSubsInS.add(sub);
                    }
                }
            }
        }

        String[] answer = new String[n];
        for (int i = 0; i < n; i++) {
            String s = arr[i];
            String bestSubForI = "";
            for (int length = 1; length <= s.length(); length++) {
                List<String> candidates = new ArrayList<>();
                for (int k = 0; k <= s.length() - length; k++) {
                    String sub = s.substring(k, k + length);
                    if (substringIndices.containsKey(sub) && substringIndices.get(sub).size() == 1) {
                        candidates.add(sub);
                    }
                }

                if (!candidates.isEmpty()) {
                    bestSubForI = candidates.stream().min(String::compareTo).get();
                    break;
                }
            }
            answer[i] = bestSubForI;
        }

        return answer;
    }
}

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Problem Info

Description

Problem 3076: Shortest Uncommon Substring in an Array - Detailed explanation and solution approaches.

Topics

Array Hash Table String Trie

Companies

AffirmAirbnbMoveworks

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:3076

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

11 steps

1
Step 1: Initialize a map (substring_indices) to store substrings as keys and sets of string indices as values. This map will track in which strings each substring appears.
2
Step 2: Iterate through each string `s` in the input array `arr` at index `i`.
3
Step 3: Generate all substrings of `s` with lengths ranging from 1 to the length of `s`.
4
Step 4: For each substring, add the index `i` to the set of indices associated with that substring in the `substring_indices` map.
5
Step 5: After processing all strings and their substrings, initialize an empty array `answer` to store the results.
6
Step 6: Iterate through each string `s` in the input array `arr` again, this time to find the shortest unique substring.
7
Step 7: For each string `s`, iterate through its substrings (again, in increasing order of length).
8
Step 8: Check if the current substring appears in the `substring_indices` map. If it does, check the size of the associated index set. If the size of the index set is 1 (meaning the substring appears only in the current string), add the substring to a list of `candidates`.
9
Step 9: If any `candidates` are found, select the lexicographically smallest substring and add it to the `answer` array. Break the inner loop to ensure that the shortest substring is chosen.
10
Step 10: If no unique substring is found for a string, add an empty string to the `answer` array.
11
Step 11: Return the `answer` array.

💡

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 * L^3)

💾

Space Complexity

O(N * L^2)

📝

Why?

Let N be the number of strings in the array and L be the maximum length of a string. Time Complexity: - Generating all substrings for a single string takes O(L^2) time because there are O(L^2) possible substrings. Checking for uniqueness involves looking it up in a hashmap which is O(1). Since we generate substrings of all lengths, we have to iterate the length of the string again O(L). - We do this for all N strings. Therefore, the time complexity for building the substring_indices map is O(N * L^2). - The second loop also iterates through each string and generates all substrings to find the shortest unique substring. This is another O(N * L^3). - Thus, the overall time complexity is O(N * L^3). Space Complexity: - The substring_indices map stores each substring and a set of string indices. In the worst case, each string has unique substrings. There can be O(L^2) substrings of length L and at most N strings so the storage space would be O(N * L^2).

📖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

Trie-based approach

A Trie can be used to store all the substrings. Each node would contain a counter of occurences. Then, iterate through substrings for each string in arr, finding the shortest one that is unique.

Time: O(N * L^2)Space: O(N * L^2)

Trade-offs: While it sounds better asymptotically, Trie implementation can be trickier than a straightforward Hash Table approach and can have larger memory overhead, particularly for small alphabets. In practice, the hash table approach is easier to code and debug, especially given the constraints.

✨

Key Insights

1Insight 1: The core idea is to efficiently identify all substrings of each string and track their occurrences across all strings in the input array.
2Insight 2: A hash table (map or dictionary) is ideal for storing substrings and the indices of the strings they appear in. This allows for quick lookup of substring occurrences.
3Insight 3: Iterating through substrings in increasing order of length allows for early termination when a unique substring is found, ensuring we find the *shortest* unique substring first. Lexicographical ordering can be done by simply using min() on a list of candidate strings of equal length.
4Insight 4: Caching already seen substrings of a string during substring generation avoids redundant calculations and insertions into the hash table, improving efficiency.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty strings in the input array. The code correctly handles this because the inner loops for substring generation will not execute, resulting in an empty string in the output for that string. Note that problem description says strings are non-empty.
Edge case 2: A string has no unique substrings. The algorithm handles this by initializing the result to an empty string. The inner loops will complete without finding any unique substrings and the output will be an empty string.
Edge case 3: Multiple shortest substrings exist. The `min()` function (or its equivalent in Java/C++) guarantees that the lexicographically smallest one will be selected.
Edge case 4: All input strings are identical. The algorithm correctly identifies that no string has a unique substring and will return an array of empty strings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible substrings. Ensure that all substrings of all possible lengths are generated and checked.

✗Common mistake 2: Not handling cases where no unique substring exists. Remember to return an empty string if no unique substring is found.

✗Common mistake 3: Not considering the lexicographical order when multiple unique substrings of the same length exist. Use the min() or String.compareTo() function.

✗Common mistake 4: Not understanding the time and space complexity implications of generating all possible substrings.

✅

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: Plagiarism detection. Finding substrings that are unique to a document can help identify original content and potential plagiarism.

🏦Real-world scenario 2: Data compression. Identifying frequently occurring substrings can be used for data compression algorithms.

🎮Real-world scenario 3: Bioinformatics. Finding unique DNA sequences within a genome can be helpful in identifying specific genes or regions.

💡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

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

⚠️ 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: June 17, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 Uncommon Substring in an Array"?

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

🤖 Need help with this problem?