Question

884. Uncommon Words from Two Sentences

中文文档

Description

A sentence is a string of single-space separated words where 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.

Given two sentences s1 and s2, return _a list of all the uncommon words_. You may return the answer in any order.

 

Example 1:

Input: s1 = "this apple is sweet", s2 = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: s1 = "apple apple", s2 = "banana"
Output: ["banana"]

 

Constraints:

• 1 <= s1.length, s2.length <= 200
• s1 and s2 consist of lowercase English letters and spaces.
• s1 and s2 do not have leading or trailing spaces.
• All the words in s1 and s2 are separated by a single space.

Solutions

Solution 1: Hash Table

According to the problem description, as long as a word appears once, it meets the requirements of the problem. Therefore, we use a hash table cnt to record all words and their occurrence counts.

Then we traverse the hash table, and take out all strings that appear only once.

The time complexity is $O(m + n)$, and the space complexity is $O(m + n)$. Here, $m$ and $n$ are the lengths of strings s1 and s2, respectively.

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]

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

class Solution {
public:
  vector<string> uncommonFromSentences(string s1, string s2) {
    unordered_map<string, int> cnt;
    auto add = [&](string& s) {
      stringstream ss(s);
      string w;
      while (ss >> w) ++cnt[move(w)];
    };
    add(s1);
    add(s2);
    vector<string> ans;
    for (auto& [s, v] : cnt)
      if (v == 1) ans.emplace_back(s);
    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;
}

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
  }
}

/**
 * @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;
};