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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.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• 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;
      }
  }
  
  ############
  
  class Solution {
      public String[] uncommonFromSentences(String s1, String s2) {
          Map<String, Integer> cnt = new HashMap<>();
          for (String s : s1.split(" ")) {
              cnt.put(s, cnt.getOrDefault(s, 0) + 1);
          }
          for (String s : s2.split(" ")) {
              cnt.put(s, cnt.getOrDefault(s, 0) + 1);
          }
          List<String> ans = new ArrayList<>();
          for (var e : cnt.entrySet()) {
              if (e.getValue() == 1) {
                  ans.add(e.getKey());
              }
          }
          return ans.toArray(new String[0]);
      }
  }
  

• // 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
  

• func uncommonFromSentences(s1 string, s2 string) (ans []string) {
  	cnt := map[string]int{}
  	for _, s := range strings.Split(s1, " ") {
  		cnt[s]++
  	}
  	for _, s := range strings.Split(s2, " ") {
  		cnt[s]++
  	}
  	for s, v := range cnt {
  		if v == 1 {
  			ans = append(ans, s)
  		}
  	}
  	return
  }
  

• function uncommonFromSentences(s1: string, s2: string): string[] {
      const cnt: Map<string, number> = new Map();
      for (const s of [...s1.split(' '), ...s2.split(' ')]) {
          cnt.set(s, (cnt.get(s) || 0) + 1);
      }
      const ans: Array<string> = [];
      for (const [s, v] of cnt.entries()) {
          if (v == 1) {
              ans.push(s);
          }
      }
      return ans;
  }
  
  

• /**
   * @param {string} s1
   * @param {string} s2
   * @return {string[]}
   */
  var uncommonFromSentences = function (s1, s2) {
      const cnt = new Map();
      for (const s of [...s1.split(' '), ...s2.split(' ')]) {
          cnt.set(s, (cnt.get(s) || 0) + 1);
      }
      const ans = [];
      for (const [s, v] of cnt.entries()) {
          if (v == 1) {
              ans.push(s);
          }
      }
      return ans;
  };
  
  

• use std::collections::HashMap;
  
  impl Solution {
      pub fn uncommon_from_sentences(s1: String, s2: String) -> Vec<String> {
          let mut map = HashMap::new();
          for s in s1.split(' ') {
              map.insert(s, !map.contains_key(s));
          }
          for s in s2.split(' ') {
              map.insert(s, !map.contains_key(s));
          }
          let mut res = Vec::new();
          for (k, v) in map {
              if v {
                  res.push(String::from(k))
              }
          }
          res
      }
  }
  
  

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