Solution Explanation: Count Common Words With One Occurrence

This problem requires finding the number of words that appear exactly once in both input arrays, words1 and words2. The most efficient approach uses hash tables (or dictionaries in Python) to count word occurrences in each array.

Algorithm:

1. Count Occurrences: Two hash tables (or dictionaries) are used: cnt1 for words1 and cnt2 for words2. Each table stores words as keys and their counts as values. We iterate through each array, updating the counts in the corresponding hash table. For example, if a word is already present, its count is incremented; otherwise, it's added with a count of 1.

2. Find Common Words with Single Occurrences: We iterate through the keys (words) of cnt1. For each word, we check two conditions:

   • The word's count in cnt1 is 1 (cnt1[word] == 1).
   • The word's count in cnt2 is also 1 (cnt2[word] == 1 or cnt2.get(word,0) == 1 in languages that handle missing keys differently).

3. Increment Count: If both conditions are true, it means the word appears exactly once in both arrays, and we increment a counter ans.

4. Return Count: Finally, we return the value of ans, representing the total number of common words with single occurrences.

Time Complexity: O(m + n), where m is the length of words1 and n is the length of words2. This is because we iterate through each array once to build the hash tables and then iterate through one hash table (which has at most the size of the smaller array) to find the common words. Hash table lookups are O(1) on average.

Space Complexity: O(m + n) in the worst case. This is because the hash tables could store all unique words from both arrays if all words are unique.

Code Examples (with detailed explanations):

Python:

from collections import Counter
 
class Solution:
    def countWords(self, words1: list[str], words2: list[str]) -> int:
        """
        Counts common words with one occurrence in two arrays.
 
        Args:
            words1: The first array of strings.
            words2: The second array of strings.
 
        Returns:
            The number of common words with exactly one occurrence in each array.
        """
 
        # Using Counter for efficient counting
        cnt1 = Counter(words1)  # Creates a dictionary with word counts for words1
        cnt2 = Counter(words2)  # Creates a dictionary with word counts for words2
 
        ans = 0  # Initialize the count of common words
        for word, count in cnt1.items():  # Iterate through words and their counts in cnt1
            if count == 1 and cnt2[word] == 1:  # Check if count is 1 in both dictionaries
                ans += 1  # Increment the count if the word appears once in both
        return ans
 

Java:

import java.util.HashMap;
import java.util.Map;
 
class Solution {
    public int countWords(String[] words1, String[] words2) {
        Map<String, Integer> cnt1 = new HashMap<>();
        Map<String, Integer> cnt2 = new HashMap<>();
 
        // Count occurrences in words1
        for (String word : words1) {
            cnt1.put(word, cnt1.getOrDefault(word, 0) + 1);
        }
 
        // Count occurrences in words2
        for (String word : words2) {
            cnt2.put(word, cnt2.getOrDefault(word, 0) + 1);
        }
 
        int ans = 0;
        for (Map.Entry<String, Integer> entry : cnt1.entrySet()) {
            String word = entry.getKey();
            int count = entry.getValue();
            if (count == 1 && cnt2.getOrDefault(word, 0) == 1) {
                ans++;
            }
        }
        return ans;
    }
}

The Java and other language examples follow a similar structure, utilizing hash map data structures to achieve the same efficient counting and comparison. The key is the use of hash tables for O(1) average-case lookup time.