# Posts published in “String”

Solve a given equation and return the value of x in the form of string “x=#value”. The equation contains only ‘+’, ‘-‘ operation, the variable x and its coefficient.

If there is no solution for the equation, return “No solution”.

If there are infinite solutions for the equation, return “Infinite solutions”.

If there is exactly one solution for the equation, we ensure that the value of x is an integer.

Example 1:

Input: "x+5-3+x=6+x-2"
Output: "x=2"


Example 2:

Input: "x=x"
Output: "Infinite solutions"


Example 3:

Input: "2x=x"
Output: "x=0"


Example 4:

Input: "2x+3x-6x=x+2"
Output: "x=-1"


Example 5:

Input: "x=x+2"
Output: "No solution"

## Solution: Parse the equation

Time complexity: O(n)
Space complexity: O(1)

## C++

We are given a personal information string S, which may represent either an email address or a phone number.

We would like to mask this personal information according to the following rules:

We define a name to be a string of length ≥ 2 consisting of only lowercase letters a-z or uppercase letters A-Z.

An email address starts with a name, followed by the symbol '@', followed by a name, followed by the dot '.' and followed by a name.

All email addresses are guaranteed to be valid and in the format of "name1@name2.name3".

To mask an email, all names must be converted to lowercase and all letters between the first and last letter of the first name must be replaced by 5 asterisks '*'.

2. Phone number:

A phone number is a string consisting of only the digits 0-9 or the characters from the set {'+', '-', '(', ')', ' '}. You may assume a phone number contains 10 to 13 digits.

The last 10 digits make up the local number, while the digits before those make up the country code. Note that the country code is optional. We want to expose only the last 4 digits and mask all other digits.

The local number should be formatted and masked as "***-***-1111", where 1 represents the exposed digits.

To mask a phone number with country code like "+111 111 111 1111", we write it in the form "+***-***-***-1111".  The '+' sign and the first '-' sign before the local number should only exist if there is a country code.  For example, a 12 digit phone number mask should start with "+**-".

Note that extraneous characters like "(", ")", " ", as well as extra dashes or plus signs not part of the above formatting scheme should be removed.

Return the correct “mask” of the information provided.

Example 1:

Input: "LeetCode@LeetCode.com"
Output: "l*****e@leetcode.com"
Explanation: All names are converted to lowercase, and the letters between the
first and last letter of the first name is replaced by 5 asterisks.
Therefore, "leetcode" -> "l*****e".


Example 2:

Input: "AB@qq.com"
Output: "a*****b@qq.com"
Explanation: There must be 5 asterisks between the first and last letter
of the first name "ab". Therefore, "ab" -> "a*****b".


Example 3:

Input: "1(234)567-890"
Output: "***-***-7890"
Explanation: 10 digits in the phone number, which means all digits make up the local number.


Example 4:

Input: "86-(10)12345678"
Output: "+**-***-***-5678"
Explanation: 12 digits, 2 digits for country code and 10 digits for local number.


Notes:

1. S.length <= 40.
2. Emails have length at least 8.
3. Phone numbers have length at least 10.

## Solution: String

Time complexity: O(n)
Space complexity: O(n)

## C++

Convert a non-negative integer to its english words representation. Given input is guaranteed to be less than 231 – 1.

Example 1:

Input: 123
Output: "One Hundred Twenty Three"


Example 2:

Input: 12345
Output: "Twelve Thousand Three Hundred Forty Five"

Example 3:

Input: 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Sixty Seven"


Example 4:

Input: 1234567891
Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Thousand Eight Hundred Ninety One"

Solution: Recursion

Time complexity: O(logn)
Space complexity: O(logn)

# Problem

https://leetcode.com/problems/reorder-log-files/description/

You have an array of logs.  Each log is a space delimited string of words.

For each log, the first word in each log is an alphanumeric identifier.  Then, either:

• Each word after the identifier will consist only of lowercase letters, or;
• Each word after the identifier will consist only of digits.

We will call these two varieties of logs letter-logs and digit-logs.  It is guaranteed that each log has at least one word after its identifier.

Reorder the logs so that all of the letter-logs come before any digit-log.  The letter-logs are ordered lexicographically ignoring identifier, with the identifier used in case of ties.  The digit-logs should be put in their original order.

Return the final order of the logs.

Example 1:

Input: ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
Output: ["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]


Note:

1. 0 <= logs.length <= 100
2. 3 <= logs[i].length <= 100
3. logs[i] is guaranteed to have an identifier, and a word after the identifier.

# Solution: Partition + Sort

1. partition the array such that all digit logs are after all letter logs
2. sort the letter logs part based on the log content

Time complexity: O(n + aloga)

Space complexity: O(n)

# Problem

https://leetcode.com/problems/replace-words/description/

In English, we have a concept called root, which can be followed by some other words to form another longer word – let’s call this word successor. For example, the root an, followed by other, which can form another word another.

Now, given a dictionary consisting of many roots and a sentence. You need to replace all the successor in the sentence with the root forming it. If a successor has many roots can form it, replace it with the root with the shortest length.

You need to output the sentence after the replacement.

Example 1:

Input: dict = ["cat", "bat", "rat"]
sentence = "the cattle was rattled by the battery"
Output: "the cat was rat by the bat"


Note:

1. The input will only have lower-case letters.
2. 1 <= dict words number <= 1000
3. 1 <= sentence words number <= 1000
4. 1 <= root length <= 100
5. 1 <= sentence words length <= 1000

# Solution 1: HashTable

Time complexity: O(sum(w^2))

Space complexity: O(sum(l))

## Solution2: Trie

Time complexity: O(sum(l) + n)

Space complexity: O(sum(l) * 26)

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