221129 :: [leetcode] Unique Email Addresses, Assign Cookies, Word Break

Unique Email Addresses

 

Every valid email consists of a local name and a domain name, separated by the '@' sign. Besides lowercase letters, the email may contain one or more '.' or '+'.

• For example, in "alice@leetcode.com", "alice" is the local name, and "leetcode.com" is the domain name.

If you add periods '.' between some characters in the local name part of an email address, mail sent there will be forwarded to the same address without dots in the local name. Note that this rule does not apply to domain names.

• For example, "alice.z@leetcode.com" and "alicez@leetcode.com" forward to the same email address.

If you add a plus '+' in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered. Note that this rule does not apply to domain names.

• For example, "m.y+name@email.com" will be forwarded to "my@email.com".

It is possible to use both of these rules at the same time.

Given an array of strings emails where we send one email to each emails[i], return the number of different addresses that actually receive mails.

 

Example 1:

Input: emails = ["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
Output: 2
Explanation: "testemail@leetcode.com" and "testemail@lee.tcode.com" actually receive mails.

 

Example 2:

Input: emails = ["a@leetcode.com","b@leetcode.com","c@leetcode.com"]
Output: 3

 

Constraints:

• 1 <= emails.length <= 100
• 1 <= emails[i].length <= 100
• emails[i] consist of lowercase English letters, '+', '.' and '@'.
• Each emails[i] contains exactly one '@' character.
• All local and domain names are non-empty.
• Local names do not start with a '+' character.
• Domain names end with the ".com" suffix.

 

실제로 메일을 받는 유효한 메일 주소의 개수 반환

 

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails = function (emails) {
    const set = new Set();
    emails.forEach((email) => {
        email = email.split("@");
        email[0] = email[0].split("+")[0].replace(/\./g, "");
        set.add(email.join("@"));
    });
    return set.size;
};

 

Assign Cookies

 

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.

Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

 

Example 1:

Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

Input: g = [1,2], s = [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

 

Constraints:

• 1 <= g.length <= 3 * 104
• 0 <= s.length <= 3 * 104
• 1 <= g[i], s[j] <= 231 - 1

 

/**
 * @param {number[]} g
 * @param {number[]} s
 * @return {number}
 */
var findContentChildren = function (g, s) {
    g.sort((a, b) => a - b);
    s.sort((a, b) => a - b);

    let i = 0;
    for (const cookie of s) {
        if (cookie >= g[i]) i++;
    }
    return i;
};

 

Word Break

 

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

 

Example 1:

Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: false

 

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

 

/**
 * @param {string} s
 * @param {string[]} wordDict
 * @return {boolean}
 */
var wordBreak = function (s, wordDict) {
    const dp = new Array(s.length + 1).fill(false).fill(true, 0, 1);
    for (let i = 0; i <= s.length; i++) {
        if (dp[i]) {
            for (let word of wordDict) {
                const newword = s.substring(i);
                if (newword.indexOf(word) === 0) {
                    dp[i + word.length] = true;
                }
            }
        }
    }
    return dp[s.length];
};