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花花酱 LeetCode 2181. Merge Nodes in Between Zeros

By zxi on March 2, 2022

You are given the head of a linked list, which contains a series of integers separated by 0‘s. The beginning and end of the linked list will have Node.val == 0.

For every two consecutive 0‘s, merge all the nodes lying in between them into a single node whose value is the sum of all the merged nodes. The modified list should not contain any 0‘s.

Return the head of the modified linked list.

Example 1:

Input: head = [0,3,1,0,4,5,2,0]
Output: [4,11]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 3 + 1 = 4.
- The sum of the nodes marked in red: 4 + 5 + 2 = 11.

Example 2:

Input: head = [0,1,0,3,0,2,2,0]
Output: [1,3,4]
Explanation: 
The above figure represents the given linked list. The modified list contains
- The sum of the nodes marked in green: 1 = 1.
- The sum of the nodes marked in red: 3 = 3.
- The sum of the nodes marked in yellow: 2 + 2 = 4.

Constraints:

  • The number of nodes in the list is in the range [3, 2 * 105].
  • 0 <= Node.val <= 1000
  • There are no two consecutive nodes with Node.val == 0.
  • The beginning and end of the linked list have Node.val == 0.

Solution: List

Skip the first zero, replace every zero node with the sum of values of its previous nodes.

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

C++

花花酱 LeetCode 2180. Count Integers With Even Digit Sum

By zxi on February 28, 2022

Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.

The digit sum of a positive integer is the sum of all its digits.

Example 1:

Input: num = 4
Output: 2
Explanation:
The only integers less than or equal to 4 whose digit sums are even are 2 and 4.

Example 2:

Input: num = 30
Output: 14
Explanation:
The 14 integers less than or equal to 30 whose digit sums are even are
2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.

Constraints:

  • 1 <= num <= 1000

Solution: Brute Force

Use std::to_string to convert an integer to string.

Time complexity: O(nlgn)
Space complexity: O(lgn)

C++

花花酱 LeetCode 2188. Minimum Time to Finish the Race

By zxi on February 28, 2022

You are given a 0-indexed 2D integer array tires where tires[i] = [fi, ri] indicates that the ith tire can finish its xth successive lap in fi * ri(x-1) seconds.

  • For example, if fi = 3 and ri = 2, then the tire would finish its 1st lap in 3 seconds, its 2nd lap in 3 * 2 = 6 seconds, its 3rd lap in 3 * 22 = 12 seconds, etc.

You are also given an integer changeTime and an integer numLaps.

The race consists of numLaps laps and you may start the race with any tire. You have an unlimited supply of each tire and after every lap, you may change to any given tire (including the current tire type) if you wait changeTime seconds.

Return the minimum time to finish the race.

Example 1:

Input: tires = [[2,3],[3,4]], changeTime = 5, numLaps = 4
Output: 21
Explanation: 
Lap 1: Start with tire 0 and finish the lap in 2 seconds.
Lap 2: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Lap 3: Change tires to a new tire 0 for 5 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Total time = 2 + 6 + 5 + 2 + 6 = 21 seconds.
The minimum time to complete the race is 21 seconds.

Example 2:

Input: tires = [[1,10],[2,2],[3,4]], changeTime = 6, numLaps = 5
Output: 25
Explanation: 
Lap 1: Start with tire 1 and finish the lap in 2 seconds.
Lap 2: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 3: Change tires to a new tire 1 for 6 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 5: Change tires to tire 0 for 6 seconds then finish the lap in another 1 second.
Total time = 2 + 4 + 6 + 2 + 4 + 6 + 1 = 25 seconds.
The minimum time to complete the race is 25 seconds.

Constraints:

  • 1 <= tires.length <= 105
  • tires[i].length == 2
  • 1 <= fi, changeTime <= 105
  • 2 <= ri <= 105
  • 1 <= numLaps <= 1000

Solution: DP

Observation: since ri >= 2, we must change tire within 20 laps, otherwise it will be slower.

pre-compute the time to finish k laps using each type of tire (k < 20), find min for each lap.

dp[i] = best[i], i < 20,
dp[i] = min{dp[i – j] + changeTime + best[j]}, i > 20

Time complexity: O(n)
Space complexity: O(n) -> O(20)

C++

花花酱 LeetCode 2187. Minimum Time to Complete Trips

By zxi on February 27, 2022

You are given an array time where time[i] denotes the time taken by the ith bus to complete one trip.

Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.

You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips.

Example 1:

Input: time = [1,2,3], totalTrips = 5
Output: 3
Explanation:
- At time t = 1, the number of trips completed by each bus are [1,0,0]. 
  The total number of trips completed is 1 + 0 + 0 = 1.
- At time t = 2, the number of trips completed by each bus are [2,1,0]. 
  The total number of trips completed is 2 + 1 + 0 = 3.
- At time t = 3, the number of trips completed by each bus are [3,1,1]. 
  The total number of trips completed is 3 + 1 + 1 = 5.
So the minimum time needed for all buses to complete at least 5 trips is 3.

Example 2:

Input: time = [2], totalTrips = 1
Output: 2
Explanation:
There is only one bus, and it will complete its first trip at t = 2.
So the minimum time needed to complete 1 trip is 2.

Constraints:

  • 1 <= time.length <= 105
  • 1 <= time[i], totalTrips <= 107

Solution: Binary Search

Find the smallest t s.t. trips >= totalTrips.

Time complexity: O(nlogm), where m ~= 1e15
Space complexity: O(1)

C++

花花酱 LeetCode 2186. Minimum Number of Steps to Make Two Strings Anagram II

By zxi on February 26, 2022

You are given two strings s and t. In one step, you can append any character to either s or t.

Return the minimum number of steps to make s and t anagrams of each other.

An anagram of a string is a string that contains the same characters with a different (or the same) ordering.

Example 1:

Input: s = "leetcode", t = "coats"
Output: 7
Explanation: 
- In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas".
- In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede".
"leetcodeas" and "coatsleede" are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

Example 2:

Input: s = "night", t = "thing"
Output: 0
Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.

Constraints:

  • 1 <= s.length, t.length <= 2 * 105
  • s and t consist of lowercase English letters.

Solution: Hashtable

Record the frequency difference of each letter.

Ans = sum(diff)

Time complexity: O(m + n)
Space complexity: O(26)

C++