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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:

## Solution 1.

• 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
}