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]
andwords2[j]
consists only of lowercase English letters.
SOLUTION:
from collections import Counter
class Solution:
def countWords(self, words1: List[str], words2: List[str]) -> int:
w1 = Counter(words1)
w2 = Counter(words2)
op = set()
for a in w1:
if w1[a] == 1 and w2[a] == 1:
op.add(a)
for a in w2:
if w1[a] == 1 and w2[a] == 1:
op.add(a)
return len(op)
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