# 884. Uncommon Words from Two Sentences

## 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

• 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) {
}
}
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)];
};
vector<string> ans;
for (auto& [s, v] : cnt)
if (v == 1) ans.emplace_back(s);
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]


• 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
}
}