What is the time complexity of Groups of Strings?

The optimal solution has a time complexity of O(N + U^2).

What data structures are used to solve Groups of Strings?

This problem uses String, Bit Manipulation, Union Find. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Groups of Strings?

Insight 1: Representing each string as a bitmask allows for efficient comparison and manipulation of character sets. Insight 2: The Union-Find data structure is ideal for grouping connected components and determining the size of each group.

Which companies ask LeetCode 2157. Groups of Strings in interviews?

This problem is frequently asked at Lowe's.

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Groups of Strings - LeetCode 2157 Solution

Groups of Strings - Complete Solution Guide

Groups of Strings is LeetCode problem 2157, 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

You are given a 0-indexed array of strings words . Each string consists of lowercase English letters only. No letter occurs more than once in any string of words . Two strings s1 and s2 are said to be connected if the set of letters of s2 can be obtained from the set of letters of s1 by any one of the following operations: Adding exactly one letter to the set of the letters of s1 . Deleting exactly one letter from the set of the letters of s1 . Replacing exactly one letter from the set of the le

Detailed Explanation

The problem requires dividing an array of strings into groups based on the connectivity between the strings. Two strings are considered connected if one can be transformed into the other by adding, deleting, or replacing a single character. The goal is to determine the maximum number of such groups and the size of the largest group.

Solution Approach

The solution uses a combination of bit manipulation and the Union-Find data structure. First, each string is converted into a bitmask where each bit represents the presence of a character (a-z). Then, the Union-Find algorithm is used to group together strings that are connected according to the problem's definition. Finally, the number of groups (number of distinct roots in Union-Find) and the size of the largest group (largest component size in Union-Find) are calculated.

Step-by-Step Algorithm

Step 1: Create a bitmask for each word in the input array. Iterate through each character in the word and set the corresponding bit in the mask to 1.
Step 2: Count occurrences of each unique bitmask. This is stored in a hash map (or similar data structure).
Step 3: Create a Union-Find data structure, where each unique bitmask is initially in its own set. The size of each set is initialized to the count of the corresponding bitmask.
Step 4: Iterate through all unique bitmasks. For each bitmask, generate all possible neighbor bitmasks by adding/removing one bit and by replacing one bit with another.
Step 5: If a neighbor bitmask exists in the set of unique bitmasks, union the corresponding sets in the Union-Find data structure.
Step 6: After processing all bitmasks, iterate through the Union-Find data structure to determine the number of connected components (groups) and the size of the largest connected component.
Step 7: Return the number of groups and the size of the largest group as the result.

Key Insights

Insight 1: Representing each string as a bitmask allows for efficient comparison and manipulation of character sets.
Insight 2: The Union-Find data structure is ideal for grouping connected components and determining the size of each group.
Insight 3: The connectivity rule implies checking for strings differing by one bit (add/delete) or two bits (replace) in their bitmask representations.

Complexity Analysis

Time Complexity: O(N + U^2)

Space Complexity: O(U)

Topics

This problem involves: String, Bit Manipulation, Union Find.

Companies

Asked at: Lowe's.

Groups of Strings | LeetCode 2157

Hard

String Bit Manipulation Union Find

Time: O(N + U^2)Space: O(U)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] groupStrings(String[] words) {
        Map<Integer, Integer> maskCounts = new HashMap<>();
        for (String word : words) {
            int mask = 0;
            for (char c : word.toCharArray()) {
                mask |= 1 << (c - 'a');
            }
            maskCounts.put(mask, maskCounts.getOrDefault(mask, 0) + 1);
        }

        List<Integer> uniqueMasks = new ArrayList<>(maskCounts.keySet());
        int n = uniqueMasks.size();
        Map<Integer, Integer> maskToIndex = new HashMap<>();
        for (int i = 0; i < n; i++) {
            maskToIndex.put(uniqueMasks.get(i), i);
        }

        int[] parent = new int[n];
        int[] size = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            size[i] = maskCounts.get(uniqueMasks.get(i));
        }

        function find = new function();

        class function{
            int find(int i) {
                if (parent[i] == i) {
                    return i;
                }
                parent[i] = find(parent[i]);
                return parent[i];
            }
        }

        function union = new function();

        class function2{
            void union(int i, int j) {
                int rootI = find.find(i);
                int rootJ = find.find(j);
                if (rootI != rootJ) {
                    if (size[rootI] < size[rootJ]) {
                        int temp = rootI;
                        rootI = rootJ;
                        rootJ = temp;
                    }
                    parent[rootJ] = rootI;
                    size[rootI] += size[rootJ];
                }
            }
        }

        function2 union2 = new function2();

        for (int i = 0; i < n; i++) {
            int mask = uniqueMasks.get(i);
            // Connection by adding or deleting one character
            // This corresponds to masks differing by one bit.
            for (int j = 0; j < 26; j++) {
                int neighbor = mask ^ (1 << j);
                if (maskToIndex.containsKey(neighbor)) {
                    union2.union(i, maskToIndex.get(neighbor));
                }
            }

            // Connection by replacing one character
            // This corresponds to one bit flipped off and another flipped on.
            for (int j = 0; j < 26; j++) {
                if ((mask >> j) % 2 == 1) { // If char j exists
                    for (int k = j + 1; k < 26; k++) {
                        if ((mask >> k) % 2 == 0) { // If char k does not exist
                            int neighbor = mask ^ (1 << j) ^ (1 << k);
                            if (maskToIndex.containsKey(neighbor)) {
                                union2.union(i, maskToIndex.get(neighbor));
                            }
                        }
                    }
                }
            }
        }

        int numGroups = 0;
        int maxSize = 0;
        for (int i = 0; i < n; i++) {
            if (parent[i] == i) {
                numGroups++;
                maxSize = Math.max(maxSize, size[i]);
            }
        }

        return new int[]{numGroups, maxSize};
    }
}

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Problem Info

Description

Problem 2157: Groups of Strings - Detailed explanation and solution approaches.

Topics

String Bit Manipulation Union Find

Companies

Lowe's

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2157

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

7 steps

1
Step 1: Create a bitmask for each word in the input array. Iterate through each character in the word and set the corresponding bit in the mask to 1.
2
Step 2: Count occurrences of each unique bitmask. This is stored in a hash map (or similar data structure).
3
Step 3: Create a Union-Find data structure, where each unique bitmask is initially in its own set. The size of each set is initialized to the count of the corresponding bitmask.
4
Step 4: Iterate through all unique bitmasks. For each bitmask, generate all possible neighbor bitmasks by adding/removing one bit and by replacing one bit with another.
5
Step 5: If a neighbor bitmask exists in the set of unique bitmasks, union the corresponding sets in the Union-Find data structure.
6
Step 6: After processing all bitmasks, iterate through the Union-Find data structure to determine the number of connected components (groups) and the size of the largest connected component.
7
Step 7: Return the number of groups and the size of the largest group as the result.

💡

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 + U^2)

💾

Space Complexity

O(U)

📝

Why?

The time complexity is determined by the conversion to bitmasks (O(N)), and the Union-Find operations including the neighborhood generation, which results in a time complexity of approximately O(U^2) where U represents the distinct bitmasks. Space is determined by the storage of distinct bitmasks and associated metadata (counts, Union-Find structures), resulting in O(U).

📖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

Graph Traversal (DFS/BFS)

An alternative approach is to create a graph where nodes are words and edges exist between connected words. Then, use Depth-First Search (DFS) or Breadth-First Search (BFS) to find connected components. Each component represents a group. This approach avoids the use of Union-Find.

Time: O(N^2) in the worst case (for checking connectivity between each pair of words) + O(N) for DFS/BFS per component, thus approximately O(N^2)Space: O(N) to store the graph and visited nodes during DFS/BFS.

Trade-offs: This approach can be simpler to understand conceptually, but it might have a higher time complexity than the Union-Find approach, particularly when the number of unique bitmasks is significantly less than the number of words. However, it could be more intuitive to implement without previous knowledge of union find.

✨

Key Insights

1Insight 1: Representing each string as a bitmask allows for efficient comparison and manipulation of character sets.
2Insight 2: The Union-Find data structure is ideal for grouping connected components and determining the size of each group.
3Insight 3: The connectivity rule implies checking for strings differing by one bit (add/delete) or two bits (replace) in their bitmask representations.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty input array. The code handles it correctly by returning [0, 0] as there are no groups and no largest group.
Edge case 2: Array with a single word. The code correctly handles it by returning [1, 1], as there's one group of size one.
Edge case 3: All words are identical. The code will correctly group them into a single large group. The count for this mask will be reflected in the size of the largest group.
Edge case 4: All words are completely disjoint (no connectivity). The code correctly identifies this by creating separate groups for each word.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the connectivity check. It is crucial to handle adding/deleting and replacing operations correctly using bit manipulation.

✗Common mistake 2: Not handling the case where multiple words map to the same bitmask. This affects the size of each group.

✗Common mistake 3: Ignoring the edge case of an empty input or a single-word input.

✗Common mistake 4: Forgetting to compress the paths during `find` operation and to union by size. It increases the runtime of the solution significantly

✅

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 web pages based on keyword overlap. The 'words' could represent sets of keywords extracted from documents, and connectivity could be defined by a threshold of shared keywords.

🏦Real-world scenario 2: Network analysis where nodes are network devices and edges represent some form of communication or dependency. Grouping devices based on connectivity patterns helps in understanding network topology and identifying critical components.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

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: November 25, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

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 "Groups of Strings"?

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

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