What is the time complexity of Similar String Groups?

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

What data structures are used to solve Similar String Groups?

This problem uses Array, Hash Table, String, Depth-First Search, Breadth-First Search, Union Find. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Similar String Groups?

Insight 1: The key observation is to efficiently determine if two strings are 'similar' according to the problem's definition. We can iterate through the strings and count the number of differing characters. If it's greater than 2, they aren't similar. Insight 2: The Union-Find data structure is perfect for grouping similar strings. Each string initially belongs to its own group. When two strings are found to be similar, we merge their groups.

Which companies ask LeetCode 839. Similar String Groups in interviews?

This problem is frequently asked at DoorDash.

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Similar String Groups | LeetCode 839

Hard

Array Hash Table String Depth-First Search Breadth-First Search Union Find

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

Solution Code

Available in:

class Solution {
    public int numSimilarGroups(String[] strs) {
        int n = strs.length;
        int[] parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
        int num_groups = n;

        function find = (i) -> {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find(parent[i]);
            return parent[i];
        };

        function union = (i, j) -> {
            int root_i = find(i);
            int root_j = find(j);
            if (root_i != root_j) {
                parent[root_i] = root_j;
                num_groups--;
            }
        };
        
        int m = strs[0].length();
        function are_similar = (s1, s2) -> {
            int diff = 0;
            for (int k = 0; k < m; k++) {
                if (s1.charAt(k) != s2.charAt(k)) {
                    diff += 1;
                    if (diff > 2) {
                        return false;
                    }
                }
            }
            return true;
        };

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (are_similar.apply(null, new String[]{strs[i], strs[j]})) {
                    union.apply(null, new Integer[]{i, j});
                }
            }
        }
        
        return num_groups;
    }

    interface function {
        Object apply(Object o, Object... args);
    }
}

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Problem Info

Description

Problem 839: Similar String Groups - Detailed explanation and solution approaches.

Topics

Array Hash Table String Depth-First Search Breadth-First Search Union Find

Companies

DoorDash

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:839

Problem Explanation

The problem asks us to find the number of 'similar string groups' within a given list of strings. Two strings are considered 'similar' if they are either identical or can be made identical by swapping at most two characters. A 'similar string group' is a set of strings where each string is similar to at least one other string in the group. The input is a list of strings `strs` where all strings are anagrams of each other. We need to return the number of such groups.

💡

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

The solution uses the Union-Find (Disjoint Set) data structure to solve this problem. The core idea is to represent each string as a node and connect two nodes (strings) if they are similar. The number of disjoint sets at the end gives the number of similar string groups. The 'are_similar' function determines if two strings are similar according to the definition in the problem. The 'find' and 'union' functions are standard Union-Find operations.

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a Union-Find data structure. Each string is initially in its own disjoint set (group). `parent[i] = i` for all strings.
2
Step 2: Iterate through all pairs of strings in the input list. For each pair, check if they are 'similar' using the `are_similar` function.
3
Step 3: If two strings are determined to be 'similar', merge their corresponding sets using the `union` function.
4
Step 4: The `find` function is used to determine the representative element (root) of the set to which a string belongs. This is used by the `union` to determine which sets should be merged. It also uses path compression to improve performance.
5
Step 5: After iterating through all pairs, the number of disjoint sets remaining is the number of 'similar string groups'. This is maintained by `num_groups` variable that gets decremented each time `union` finds two different roots and merges them.
6
Step 6: Return the final `num_groups`.

💡

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 * m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the nested loop comparing all pairs of strings. The are_similar function contributes a factor of m to this complexity. The space complexity is determined by the size of the parent array which stores the root of each string in the union find data structure.

📖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

Depth-First Search (DFS)

Represent the strings as nodes in a graph. Two nodes are connected if the corresponding strings are similar. Then, perform a DFS to find the connected components. Each connected component represents a similar string group.

Time: O(n^2 * m), where n is the number of strings and m is the length of each string.Space: O(n^2), in worst case where all strings are similar creating an adjacency matrix.

Trade-offs: The DFS approach is conceptually similar to Union-Find. While it achieves the same time complexity for the similarity check and traversal of all strings pairs, the space complexity could be O(n^2) if implemented via an adjacency matrix, making Union-Find preferable. DFS can also be more prone to stack overflow errors for large n.

✨

Key Insights

1Insight 1: The key observation is to efficiently determine if two strings are 'similar' according to the problem's definition. We can iterate through the strings and count the number of differing characters. If it's greater than 2, they aren't similar.
2Insight 2: The Union-Find data structure is perfect for grouping similar strings. Each string initially belongs to its own group. When two strings are found to be similar, we merge their groups.
3Insight 3: Since the strings are anagrams of each other, we don't need to consider cases where strings of different lengths are present. The problem also constrains the length of each string, influencing the runtime.

📚

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: Input with only one string. In this case, the answer is 1, as the string forms a group by itself. The algorithm correctly handles this case because the outer loops don't execute.
Edge case 2: All strings are identical. In this case, all strings belong to the same group, and the answer is 1. The algorithm handles this because the `are_similar` function will always return true, and all sets will be merged into one.
Edge case 3: No similar strings exist. In this case, each string forms its own group, and the answer is the number of strings. The algorithm handles this because the `union` function will never be called.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not correctly implementing the Union-Find data structure, especially the `find` function with path compression. This can lead to significantly worse performance.

✗Common mistake 2: Incorrectly determining if two strings are 'similar'. Forgetting that strings can be identical or only requiring a *single* swap and not handling that edge case.

✗Common mistake 3: Not considering the constraints given. For instance, assuming strings might have different lengths and adding checks for that.

✅

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: Clustering similar documents or articles. The strings could represent the content of documents, and similarity could be defined based on the number of common words or phrases. This can be used for information retrieval and recommendation systems.

🏦Real-world scenario 2: Network analysis. Strings could represent network nodes or entities, and similarity could be defined based on shared connections or properties. Finding similar string groups could help identify communities or clusters within the network.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 28, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 29

❓

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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Similar String Groups"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?

Similar String Groups - Complete Solution Guide

Problem Statement

Detailed Explanation

Solution Approach

Step-by-Step Algorithm

Key Insights

Complexity Analysis

Topics

Companies

Similar String Groups | LeetCode 839

Solution Code

Problem Info

Description

Topics

Companies

Stats

Problem Explanation

Think Before You Code

🤔 Ask Yourself

✏️ Try This First

Solution Approach

Step-by-Step Solution

Pro Tip for Students

Complexity Analysis

Time Complexity

Space Complexity

Why?

Big O Notation - Quick Guide

Alternative Approaches

Depth-First Search (DFS)

Key Insights

Key Terms Explained

Array

Hash Table

String

Edge Cases

Common Mistakes to Avoid

Learning Tip

Where is This Used in Real Life?

Author's Notes

About the Author

Community Tips & Insights

Frequently Asked Questions

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

How do I choose between a hash map and a hash set?

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

What is the importance of time and space complexity analysis?

What is the optimal approach for "Similar String Groups"?

Similar String Groups | LeetCode 839

Solution Code

Problem Info

Description

Topics

Companies

Stats

Problem Explanation

Think Before You Code

🤔 Ask Yourself

✏️ Try This First

Solution Approach

Step-by-Step Solution

Pro Tip for Students

Complexity Analysis

Time Complexity

Space Complexity

Why?

Big O Notation - Quick Guide

Alternative Approaches

Depth-First Search (DFS)

Key Insights

Key Terms Explained

Array

Hash Table

String

Edge Cases

Common Mistakes to Avoid

Learning Tip

Where is This Used in Real Life?

Author's Notes

About the Author