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:

## Solution 1.

// OJ: https://leetcode.com/problems/count-common-words-with-one-occurrence/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int countWords(vector<string>& A, vector<string>& B) {
unordered_map<string, int> a, b;
for (auto &s : A) a[s]++;
for (auto &s : B) b[s]++;
int ans = 0;
for (auto &[s, cnt] : a) {
if (cnt == 1 && b.count(s) && b[s] == 1) ++ans;
}
return ans;
}
};