What is the time complexity of Count Pairs Of Similar Strings?

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

What data structures are used to solve Count Pairs Of Similar Strings?

This problem uses Array, Hash Table, String, Bit Manipulation, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Pairs Of Similar Strings?

Insight 1: Using sets to efficiently determine if two strings have the same characters. Sets only store unique elements, making comparison straightforward. Insight 2: Nested loops are necessary to compare all possible pairs of strings. This results in a time complexity dependent on the square of the number of strings.

Which companies ask LeetCode 2506. Count Pairs Of Similar Strings in interviews?

This problem is frequently asked at IBM.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Count Pairs Of Similar Strings - LeetCode 2506 Solution

Q: How to solve Count Pairs Of Similar Strings?

The provided solutions all follow a brute-force approach. They use nested loops to iterate through all possible pairs of strings. For each pair, they determine if the strings are similar by comparing the sets of unique characters present in each string. If the sets are identical, the pair count is incremented. The Python and Java solutions use built-in set functionality, while the C++ and C solutions manually create and compare character sets.

Count Pairs Of Similar Strings - Complete Solution Guide

Count Pairs Of Similar Strings is LeetCode problem 2506, a Easy 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 string array words . Two strings are similar if they consist of the same characters. For example, "abca" and "cba" are similar since both consist of characters 'a' , 'b' , and 'c' . However, "abacba" and "bcfd" are not similar since they do not consist of the same characters. Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar . Example 1: Input: words = ["aba","aabb","abcd","bac","aabc"] Output

Detailed Explanation

The problem asks you to count the number of pairs of strings in a given array where each pair consists of two strings that contain the same set of characters. The order of characters within a string doesn't matter; only the unique characters present are significant. For example, "abc" and "bca" are considered similar, while "abc" and "abd" are not. The problem explicitly states that you only need to consider pairs (i, j) where i < j, ensuring you don't double-count pairs.

Solution Approach

The provided solutions all follow a brute-force approach. They use nested loops to iterate through all possible pairs of strings. For each pair, they determine if the strings are similar by comparing the sets of unique characters present in each string. If the sets are identical, the pair count is incremented. The Python and Java solutions use built-in set functionality, while the C++ and C solutions manually create and compare character sets.

Step-by-Step Algorithm

Step 1: Initialize a count variable to 0.
Step 2: Use nested loops to iterate through all pairs (i, j) of strings in the input array, where i < j.
Step 3: For each pair, create sets (or equivalent arrays) to store the unique characters of each string.
Step 4: Compare the two sets (or arrays). If they are equal (containing the same unique characters), increment the count.
Step 5: After iterating through all pairs, return the final count.

Key Insights

Insight 1: Using sets to efficiently determine if two strings have the same characters. Sets only store unique elements, making comparison straightforward.
Insight 2: Nested loops are necessary to compare all possible pairs of strings. This results in a time complexity dependent on the square of the number of strings.
Insight 3: The problem avoids unnecessary comparisons by iterating only through pairs (i, j) where i < j. This prevents duplicate comparisons.

Complexity Analysis

Time Complexity: O(n^2*m)

Space Complexity: O(m)

Topics

This problem involves: Array, Hash Table, String, Bit Manipulation, Counting.

Companies

Asked at: IBM.

Count Pairs Of Similar Strings | LeetCode 2506

Easy

Array Hash Table String Bit Manipulation Counting

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

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int similarPairs(String[] words) {
        int count = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (areSimilar(words[i], words[j])) {
                    count++;
                }
            }
        }
        return count;
    }

    private boolean areSimilar(String word1, String word2) {
        Set<Character> set1 = new HashSet<>();
        Set<Character> set2 = new HashSet<>();

        for (char c : word1.toCharArray()) {
            set1.add(c);
        }
        for (char c : word2.toCharArray()) {
            set2.add(c);
        }

        return set1.equals(set2);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2506: Count Pairs Of Similar Strings - Detailed explanation and solution approaches.

Topics

Array Hash Table String Bit Manipulation Counting

Companies

IBM

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2506

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

5 steps

1
Step 1: Initialize a count variable to 0.
2
Step 2: Use nested loops to iterate through all pairs (i, j) of strings in the input array, where i < j.
3
Step 3: For each pair, create sets (or equivalent arrays) to store the unique characters of each string.
4
Step 4: Compare the two sets (or arrays). If they are equal (containing the same unique characters), increment the count.
5
Step 5: After iterating through all pairs, return the final count.

💡

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(m)

📝

Why?

The nested loops dominate the time complexity, resulting in O(n^2) where n is the number of strings. Within each iteration of the inner loop, creating the sets of unique characters takes O(m) time, where m is the maximum length of a string. The space complexity is O(m) because the size of each set is at most m (the number of unique characters in a string). The space usage is independent of the number of strings (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

Optimized Brute Force with bit manipulation

Instead of using sets, represent each string as a bitmask (integer) where each bit corresponds to a letter of the alphabet. A bit is 1 if that letter is present in the string, and 0 otherwise. Comparing the bitmasks allows for constant time similarity checks.

Time: O(n^2 + nm)Space: O(1)

Trade-offs: This approach uses slightly less space. The time complexity improvement is marginal but could be noticeable for very large strings or limited resources. The added complexity of implementing the bit manipulation might not be worth it unless string lengths are very large and memory is a significant constraint.

✨

Key Insights

1Insight 1: Using sets to efficiently determine if two strings have the same characters. Sets only store unique elements, making comparison straightforward.
2Insight 2: Nested loops are necessary to compare all possible pairs of strings. This results in a time complexity dependent on the square of the number of strings.
3Insight 3: The problem avoids unnecessary comparisons by iterating only through pairs (i, j) where i < j. This prevents duplicate comparisons.

📚

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 input array. The solution correctly handles this by returning 0 since there are no pairs.
Edge case 2: Strings with all the same characters. The solution will correctly identify them as similar.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider only pairs (i, j) where i < j, leading to double-counting of similar pairs. The correct implementation ensures each pair is considered only once.

✗Common mistake 2: Not handling edge cases correctly, especially the empty input array. A robust solution should gracefully handle an empty input.

✅

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: Document similarity analysis. This algorithm can be used as a basic component to determine the similarity of documents based on the vocabulary they use.

🏦Real-world scenario 2: Data cleaning and deduplication. Identifying and removing duplicate data entries based on their constituent elements is another practical application.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: November 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Count Pairs Of Similar Strings"?

For this easy-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?