What is the time complexity of Count Common Words With One Occurrence?

The optimal solution has a time complexity of O(n + m).

What is the space complexity of Count Common Words With One Occurrence?

The space complexity is O(n + m).

What data structures are used to solve Count Common Words With One Occurrence?

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

What are the key insights for solving Count Common Words With One Occurrence?

Insight 1: Using hash tables (dictionaries in Python) to efficiently count the occurrences of each word in both arrays is crucial for achieving linear time complexity. Insight 2: Iterating through the frequency map of one array and checking for the existence and frequency of the same word in the other array's frequency map is the core algorithmic strategy.

Which companies ask LeetCode 2085. Count Common Words With One Occurrence in interviews?

This problem is frequently asked at Jane Street.

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Count Common Words With One Occurrence - LeetCode 2085 Solution

Count Common Words With One Occurrence - Complete Solution Guide

Count Common Words With One Occurrence is LeetCode problem 2085, 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

Given two string arrays words1 and words2 , return the number of strings that appear exactly once in each of the two arrays. Example 1: Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"] Output: 2 Explanation: - "leetcode" appears exactly once in each of the two arrays. We count this string. - "amazing" appears exactly once in each of the two arrays. We count this string. - "is" appears in each of the two arrays, but there are 2 occurrences of it in words

Detailed Explanation

The problem asks to find the number of strings that appear exactly once in both input string arrays, `words1` and `words2`. The input consists of two arrays of strings. The output is a single integer representing the count of common strings that appear exactly once in each array. Constraints limit the lengths of the arrays and individual strings.

Solution Approach

The provided solutions utilize hash tables (dictionaries, maps, unordered_maps) to store the frequency of each word in `words1` and `words2`. First, they count the occurrences of each word in each array separately. Then, they iterate through the frequency map of `words1`. For each word, they check if it exists in `words2`'s frequency map and if its frequency is 1 in both maps. If both conditions are true, the counter is incremented. Finally, the counter (the number of common words appearing exactly once in each array) is returned.

Step-by-Step Algorithm

Step 1: Create two hash tables (dictionaries/maps) to store word frequencies for `words1` and `words2`.
Step 2: Iterate through `words1`, updating the frequency count for each word in the `words1` frequency map.
Step 3: Iterate through `words2`, updating the frequency count for each word in the `words2` frequency map.
Step 4: Iterate through the keys (words) of the `words1` frequency map.
Step 5: For each word, check if its frequency is 1 in `words1`'s map and if it exists in `words2`'s map with a frequency of 1.
Step 6: If both conditions are true, increment the common word counter.
Step 7: Return the final count of common words.

Key Insights

Insight 1: Using hash tables (dictionaries in Python) to efficiently count the occurrences of each word in both arrays is crucial for achieving linear time complexity.
Insight 2: Iterating through the frequency map of one array and checking for the existence and frequency of the same word in the other array's frequency map is the core algorithmic strategy.
Insight 3: Handling the case where a word might not exist in one or both arrays is important to avoid errors. The `getOrDefault` method in Java and similar approaches in other languages elegantly handle this.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

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

Companies

Asked at: Jane Street.

Count Common Words With One Occurrence | LeetCode 2085

Easy

Array Hash Table String Counting

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int countWords(String[] words1, String[] words2) {
        Map<String, Integer> count1 = new HashMap<>();
        Map<String, Integer> count2 = new HashMap<>();

        for (String word : words1) {
            count1.put(word, count1.getOrDefault(word, 0) + 1);
        }

        for (String word : words2) {
            count2.put(word, count2.getOrDefault(word, 0) + 1);
        }

        int result = 0;
        for (String word : count1.keySet()) {
            if (count1.get(word) == 1 && count2.containsKey(word) && count2.get(word) == 1) {
                result++;
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 2085: Count Common Words With One Occurrence - Detailed explanation and solution approaches.

Topics

Array Hash Table String Counting

Companies

Jane Street

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2085

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 two hash tables (dictionaries/maps) to store word frequencies for `words1` and `words2`.
2
Step 2: Iterate through `words1`, updating the frequency count for each word in the `words1` frequency map.
3
Step 3: Iterate through `words2`, updating the frequency count for each word in the `words2` frequency map.
4
Step 4: Iterate through the keys (words) of the `words1` frequency map.
5
Step 5: For each word, check if its frequency is 1 in `words1`'s map and if it exists in `words2`'s map with a frequency of 1.
6
Step 6: If both conditions are true, increment the common word counter.
7
Step 7: Return the final count of common 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 + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is linear because iterating through each array and each hash table takes linear time with respect to the size of the arrays (m and n). Building the hash tables also takes linear time. The space complexity is linear because the size of the hash tables is at most proportional to the number of unique words in each array. In the worst-case scenario, all words are unique.

📖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

Nested Loops Approach

Use nested loops to compare each word in `words1` with each word in `words2`. Count occurrences separately for each word in each array. This is less efficient.

Time: O(m*n)Space: O(1)

Trade-offs: This approach is significantly slower than the hash table approach, especially for large input arrays. It has lower space complexity, but this advantage is negligible compared to the performance penalty.

✨

Key Insights

1Insight 1: Using hash tables (dictionaries in Python) to efficiently count the occurrences of each word in both arrays is crucial for achieving linear time complexity.
2Insight 2: Iterating through the frequency map of one array and checking for the existence and frequency of the same word in the other array's frequency map is the core algorithmic strategy.
3Insight 3: Handling the case where a word might not exist in one or both arrays is important to avoid errors. The `getOrDefault` method in Java and similar approaches in other languages elegantly handle this.

📚

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 arrays. The solution handles this gracefully because empty maps will not increment the counter.
Edge case 2: Arrays with no common words. The solution correctly returns 0 in this case because the condition in step 5 will never be met.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling word frequencies. Failing to check if the word exists in both frequency maps and has a frequency of 1 in both will lead to incorrect results.

✗Common mistake 2: Not considering the case of empty input arrays or arrays with no common words. Robust solutions should handle these cases explicitly.

✅

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: Analyzing website logs to find common terms used by specific user groups that are unique to those groups.

🏦Real-world scenario 2: Comparing two lists of product identifiers to find products that are exclusively listed by both vendors, and sold only once by each vendor.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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.

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Count Common Words With One Occurrence"?

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.

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