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: Hashtable
Time complexity: O(n + m)
Space complexity: O(n + m)
C++
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// Author: Huahua class Solution { public: int countWords(vector<string>& words1, vector<string>& words2) { unordered_map<string, int> c1; unordered_map<string, int> c2; for (const string& w : words1) ++c1[w]; for (const string& w : words2) ++c2[w]; int ans = 0; for (const auto& [w, c] : c1) if (c == 1 && c2[w] == 1) ++ans; return ans; } }; |
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