How do you solve Count Common Words With One Occurrence on LeetCode?

Count Common Words With One Occurrence (LeetCode #2085) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: We can check each word in the first array against every word in the second array. This is straightforward but inefficient for larger inputs.

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

The optimal solution for Count Common Words With One Occurrence runs in O(n²) time and uses O(1) space. This complexity arises because we are checking each word in words1 against every word in words2, leading to a quadratic number of checks.

What algorithm or data structure is used to solve Count Common Words With One Occurrence?

Count Common Words With One Occurrence uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Common Words With One Occurrence?

Explain your thought process clearly. Discuss the trade-offs of different approaches. Practice coding hash map solutions.

What are common mistakes when solving Count Common Words With One Occurrence?

Not using hash maps for counting occurrences. Overlooking words that appear multiple times in one array.

Count Common Words With One Occurrence — LeetCode #2085 (Easy)

Tags: Array, Hash Table, String, Counting

Related patterns: Hash Map, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

We can check each word in the first array against every word in the second array. This is straightforward but inefficient for larger inputs.

This complexity arises because we are checking each word in words1 against every word in words2, leading to a quadratic number of checks.

Step 1: For each word in words1, count its occurrences.
Step 2: For each word in words2, count its occurrences.
Step 3: Check each word from words1 against words2 to see if it appears exactly once in both.

1. words1: ['leetcode', 'is', 'amazing', 'as', 'is'] -> count1: {'leetcode': 1, 'is': 2, 'amazing': 1, 'as': 1} 2. words2: ['amazing', 'leetcode', 'is'] -> count2: {'amazing': 1, 'leetcode': 1, 'is': 1} 3. Check 'leetcode': count1[leetcode] = 1, count2[leetcode] = 1 -> count = 1 4. Check 'is': count1[is] = 2, count2[is] = 1 -> count = 1 5. Check 'amazing': count1[amazing] = 1, count2[amazing] = 1 -> count = 2 6. Check 'as': count1[as] = 1, count2[as] = 0 -> count = 2 Final count = 2.

Common Mistakes

Not using hash maps for counting occurrences.
Overlooking words that appear multiple times in one array.

Interview Tips

Explain your thought process clearly.
Discuss the trade-offs of different approaches.
Practice coding hash map solutions.

#2085

Count Common Words With One Occurrence

Easy

Array↗Hash Table↗String↗Counting↗Hash Map↗Array↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

We can check each word in the first array against every word in the second array. This is straightforward but inefficient for larger inputs.

⚙️

Algorithm

3 steps

1Step 1: For each word in words1, count its occurrences.
2Step 2: For each word in words2, count its occurrences.
3Step 3: Check each word from words1 against words2 to see if it appears exactly once in both.

solution.py14 lines

1# Full working Python code
2from collections import Counter
3
4def count_common_words(words1, words2):
5    count1 = Counter(words1)
6    count2 = Counter(words2)
7    common_count = 0
8    for word in count1:
9        if count1[word] == 1 and count2[word] == 1:
10            common_count += 1
11    return common_count
12
13# Example usage
14print(count_common_words(["leetcode","is","amazing","as","is"], ["amazing","leetcode","is"]))

ℹ

Complexity note: This complexity arises because we are checking each word in words1 against every word in words2, leading to a quadratic number of checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.