Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/884.html

884. Uncommon Words from Two Sentences (Easy)

We are given two sentences A and B.  (A sentence is a string of space separated words.  Each word consists only of lowercase letters.)

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Return a list of all uncommon words. 

You may return the list in any order.

 

Example 1:

Input: A = "this apple is sweet", B = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: A = "apple apple", B = "banana"
Output: ["banana"]

 

Note:

  1. 0 <= A.length <= 200
  2. 0 <= B.length <= 200
  3. A and B both contain only spaces and lowercase letters.

Related Topics:
Hash Table

Solution 1.

// OJ: https://leetcode.com/problems/uncommon-words-from-two-sentences/
// Time: O(A+B)
// Space: O(A+B)
class Solution {
private:
    unordered_map<string, int> getCounts(string s) {
        istringstream ss(s);
        string word;
        unordered_map<string, int> m;
        while (ss >> word) m[word]++;
        return m;
    }
public:
    vector<string> uncommonFromSentences(string A, string B) {
        auto m = getCounts(A), n = getCounts(B);
        vector<string> ans;
        for (auto &p : m) {
            if (p.second == 1 && n.find(p.first) == n.end()) ans.push_back(p.first);
        }
        for (auto &p : n) {
            if (p.second == 1 && m.find(p.first) == m.end()) ans.push_back(p.first);
        }
        return ans;
    }
};

Java

  • class Solution {
        public String[] uncommonFromSentences(String A, String B) {
            List<String> uncommonWords = new ArrayList<String>();
            Set<String> set = new HashSet<String>();
            String[] strA = A.split(" ");
            String[] strB = B.split(" ");
            for (String word : strA) {
                if (set.add(word))
                    uncommonWords.add(word);
                else
                    uncommonWords.remove(word);
            }
            for (String word : strB) {
                if (set.add(word))
                    uncommonWords.add(word);
                else
                    uncommonWords.remove(word);
            }
            int length = uncommonWords.size();
            String[] uncommonWordsArray = new String[length];
            for (int i = 0; i < length; i++)
                uncommonWordsArray[i] = uncommonWords.get(i);
            return uncommonWordsArray;
        }
    }
    
  • // OJ: https://leetcode.com/problems/uncommon-words-from-two-sentences/
    // Time: O(A+B)
    // Space: O(A+B)
    class Solution {
    private:
        unordered_map<string, int> getCounts(string s) {
            istringstream ss(s);
            string word;
            unordered_map<string, int> m;
            while (ss >> word) m[word]++;
            return m;
        }
    public:
        vector<string> uncommonFromSentences(string A, string B) {
            auto m = getCounts(A), n = getCounts(B);
            vector<string> ans;
            for (auto &p : m) {
                if (p.second == 1 && n.find(p.first) == n.end()) ans.push_back(p.first);
            }
            for (auto &p : n) {
                if (p.second == 1 && m.find(p.first) == m.end()) ans.push_back(p.first);
            }
            return ans;
        }
    };
    
  • class Solution:
        def uncommonFromSentences(self, s1: str, s2: str) -> List[str]:
            cnt = Counter(s1.split()) + Counter(s2.split())
            return [s for s, v in cnt.items() if v == 1]
    
    ############
    3
    class Solution:
        def uncommonFromSentences(self, A, B):
            """
            :type A: str
            :type B: str
            :rtype: List[str]
            """
            count_A = collections.Counter(A.split(' '))
            count_B = collections.Counter(B.split(' '))
            words = list((count_A.keys() | count_B.keys()) - (count_A.keys() & count_B.keys()))
            ans = []
            for word in words:
                if count_A[word] == 1 or count_B[word] == 1:
                    ans.append(word)
            return ans
    

All Problems

All Solutions