Coffee Time

You are given a 0-indexed 2D integer array flowers, where flowers[i] = [starti, endi] means the ith flower will be in 
full bloom from starti to endi (inclusive). You are also given a 0-indexed integer array persons of size n, 
where persons[i] is the time that the ith person will arrive to see the flowers.

Return an integer array answer of size n, where answer[i] is the number of flowers that are in full bloom when 
the ith person arrives.

Example 1:

Input: flowers = [[1,6],[3,7],[9,12],[4,13]], persons = [2,3,7,11]
Output: [1,2,2,2]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.
Example 2:

Input: flowers = [[1,10],[3,3]], persons = [3,3,2]
Output: [2,2,1]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.
 

Constraints:

1 <= flowers.length <= 5 * 104
flowers[i].length == 2
1 <= starti <= endi <= 109
1 <= persons.length <= 5 * 104
1 <= persons[i] <= 109

Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.

Note that after backspacing an empty text, the text will continue empty.

 

Example 1:

Input: s = "ab#c", t = "ad#c"
Output: true
Explanation: Both s and t become "ac".
Example 2:

Input: s = "ab##", t = "c#d#"
Output: true
Explanation: Both s and t become "".
Example 3:

Input: s = "a#c", t = "b"
Output: false
Explanation: s becomes "c" while t becomes "b".
 

Constraints:

1 <= s.length, t.length <= 200
s and t only contain lowercase letters and '#' characters.
 

Follow up: Can you solve it in O(n) time and O(1) space?

You are given a string s consisting of digits and an integer k.

A round can be completed if the length of s is greater than k. In one round, do the following:

Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.
Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.
Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.
Return s after all rounds have been completed.

 

Example 1:

Input: s = "11111222223", k = 3
Output: "135"
Explanation: 
- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".
  Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
  So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.
- For the second round, we divide s into "346" and "5".
  Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
  So, s becomes "13" + "5" = "135" after second round. 
Now, s.length <= k, so we return "135" as the answer.
Example 2:

Input: s = "00000000", k = 3
Output: "000"
Explanation: 
We divide s into "000", "000", and "00".
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".
 

Constraints:

1 <= s.length <= 100
2 <= k <= 100
s consists of digits only.

You are given two strings current and correct representing two 24-hour times.

24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.

In one operation you can increase the time current by 1, 5, 15, or 60 minutes. You can perform this operation any number of times.

Return the minimum number of operations needed to convert current to correct.

 

Example 1:

Input: current = "02:30", correct = "04:35"
Output: 3
Explanation:
We can convert current to correct in 3 operations as follows:
- Add 60 minutes to current. current becomes "03:30".
- Add 60 minutes to current. current becomes "04:30".
- Add 5 minutes to current. current becomes "04:35".
It can be proven that it is not possible to convert current to correct in fewer than 3 operations.
Example 2:

Input: current = "11:00", correct = "11:01"
Output: 1
Explanation: We only have to add one minute to current, so the minimum number of operations needed is 1.
 

Constraints:

current and correct are in the format "HH:MM"
current <= correct

You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).

Return the largest possible value of num after any number of swaps.

 

Example 1:

Input: num = 1234
Output: 3412
Explanation: Swap the digit 3 with the digit 1, this results in the number 3214.
Swap the digit 2 with the digit 4, this results in the number 3412.
Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.
Example 2:

Input: num = 65875
Output: 87655
Explanation: Swap the digit 8 with the digit 6, this results in the number 85675.
Swap the first digit 5 with the digit 7, this results in the number 87655.
Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.
 

Constraints:

1 <= num <= 10^9