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Formatted question description: https://leetcode.ca/all/2085.html

2085. Count Common Words With One Occurrence (Easy)

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 words1. We do not count this string.
- "as" appears once in words1, but does not appear in words2. We do not count this string.
Thus, there are 2 strings that appear exactly once in each of the two arrays.

Example 2:

Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
Output: 0
Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"]
Output: 1
Explanation: The only string that appears exactly once in each of the two arrays is "ab".

 

Constraints:

• 1 <= words1.length, words2.length <= 1000
• 1 <= words1[i].length, words2[j].length <= 30
• words1[i] and words2[j] consists only of lowercase English letters.

Similar Questions:

• Intersection of Two Arrays (Easy)
• Uncommon Words from Two Sentences (Easy)
• Kth Distinct String in an Array (Easy)

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public int countWords(String[] words1, String[] words2) {
          Map<String, Integer> cnt1 = count(words1);
          Map<String, Integer> cnt2 = count(words2);
          int ans = 0;
          for (String w : words1) {
              if (cnt1.getOrDefault(w, 0) == 1 && cnt2.getOrDefault(w, 0) == 1) {
                  ++ans;
              }
          }
          return ans;
      }
  
      private Map<String, Integer> count(String[] words) {
          Map<String, Integer> cnt = new HashMap<>();
          for (String w : words) {
              cnt.put(w, cnt.getOrDefault(w, 0) + 1);
          }
          return cnt;
      }
  }
  

• class Solution {
  public:
      int countWords(vector<string>& words1, vector<string>& words2) {
          unordered_map<string, int> cnt1;
          unordered_map<string, int> cnt2;
          for (auto& w : words1) cnt1[w]++;
          for (auto& w : words2) cnt2[w]++;
          int ans = 0;
          for (auto& w : words1) ans += (cnt1[w] == 1 && cnt2[w] == 1);
          return ans;
      }
  };
  

• class Solution:
      def countWords(self, words1: List[str], words2: List[str]) -> int:
          cnt1 = Counter(words1)
          cnt2 = Counter(words2)
          return sum(cnt2[k] == 1 for k, v in cnt1.items() if v == 1)
  
  

• func countWords(words1 []string, words2 []string) int {
  	cnt1 := map[string]int{}
  	cnt2 := map[string]int{}
  	for _, w := range words1 {
  		cnt1[w]++
  	}
  	for _, w := range words2 {
  		cnt2[w]++
  	}
  	ans := 0
  	for _, w := range words1 {
  		if cnt1[w] == 1 && cnt2[w] == 1 {
  			ans++
  		}
  	}
  	return ans
  }
  

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