README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0800-0899/0884.Uncommon%20Words%20from%20Two%20Sentences/README_EN.md	Hash Table String

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.

Python3

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]

Java

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]);
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

JavaScript

/**
 * @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;
};