# 2085. Count Common Words With One Occurrence

## Description

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.

## Solutions

Solution 1: Hash Table + Counting

We can use two hash tables, $cnt1$ and $cnt2$, to count the occurrences of each string in the two string arrays respectively. Then, we traverse one of the hash tables. If a string appears once in the other hash table and also appears once in the current hash table, we increment the answer by one.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Where $n$ and $m$ are the lengths of the two string arrays respectively.

• class Solution {
public int countWords(String[] words1, String[] words2) {
Map<String, Integer> cnt1 = new HashMap<>();
Map<String, Integer> cnt2 = new HashMap<>();
for (var w : words1) {
cnt1.merge(w, 1, Integer::sum);
}
for (var w : words2) {
cnt2.merge(w, 1, Integer::sum);
}
int ans = 0;
for (var e : cnt1.entrySet()) {
if (e.getValue() == 1 && cnt2.getOrDefault(e.getKey(), 0) == 1) {
++ans;
}
}
return ans;
}
}

• 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, v] : cnt1) {
ans += v == 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(v == 1 and cnt2[w] == 1 for w, v in cnt1.items())


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

• function countWords(words1: string[], words2: string[]): number {
const cnt1 = new Map<string, number>();
const cnt2 = new Map<string, number>();
for (const w of words1) {
cnt1.set(w, (cnt1.get(w) ?? 0) + 1);
}
for (const w of words2) {
cnt2.set(w, (cnt2.get(w) ?? 0) + 1);
}
let ans = 0;
for (const [w, v] of cnt1) {
if (v === 1 && cnt2.get(w) === 1) {
++ans;
}
}
return ans;
}