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花花酱 LeetCode 1335. Minimum Difficulty of a Job Schedule

You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the i-th job, you have to finish all the jobs j where 0 <= j < i).

You have to finish at least one task every day. The difficulty of a job schedule is the sum of difficulties of each day of the d days. The difficulty of a day is the maximum difficulty of a job done in that day.

Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th job is jobDifficulty[i].

Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.

Example 1:

Input: jobDifficulty = [6,5,4,3,2,1], d = 2
Output: 7
Explanation: First day you can finish the first 5 jobs, total difficulty = 6.
Second day you can finish the last job, total difficulty = 1.
The difficulty of the schedule = 6 + 1 = 7 

Example 2:

Input: jobDifficulty = [9,9,9], d = 4
Output: -1
Explanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.

Example 3:

Input: jobDifficulty = [1,1,1], d = 3
Output: 3
Explanation: The schedule is one job per day. total difficulty will be 3.

Example 4:

Input: jobDifficulty = [7,1,7,1,7,1], d = 3
Output: 15

Example 5:

Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6
Output: 843

Constraints:

  • 1 <= jobDifficulty.length <= 300
  • 0 <= jobDifficulty[i] <= 1000
  • 1 <= d <= 10

Solution: DP

dp[i][k] := min difficulties to schedule jobs 1~i in k days.

Schedule 1 ~ j in k – 1 days and schedule j + 1 ~ i in 1 day.

Init: dp[0][0] = 0
Transition: dp[i][k] := min(dp[j][k -1] + max(jobs[j + 1 ~ i]), k – 1 <= j < i
Answer: dp[n][d]

Time complexity: O(n^2*d)
Space complexity: O(n)

C++

花花酱 LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.

Return the city with the smallest numberof cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.

Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.

Example 1:

Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2] 
City 1 -> [City 0, City 2, City 3] 
City 2 -> [City 0, City 1, City 3] 
City 3 -> [City 1, City 2] 
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1] 
City 1 -> [City 0, City 4] 
City 2 -> [City 3, City 4] 
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3] 
The city 0 has 1 neighboring city at a distanceThreshold = 2.

Constraints:

  • 2 <= n <= 100
  • 1 <= edges.length <= n * (n - 1) / 2
  • edges[i].length == 3
  • 0 <= fromi < toi < n
  • 1 <= weighti, distanceThreshold <= 10^4
  • All pairs (fromi, toi) are distinct.

Solution: Floyd-Warshall

All pair shortest path

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

花花酱 LeetCode 1333. Filter Restaurants by Vegan-Friendly, Price and Distance

Given the array restaurants where  restaurants[i] = [idi, ratingi, veganFriendlyi, pricei, distancei]. You have to filter the restaurants using three filters.

The veganFriendly filter will be either true (meaning you should only include restaurants with veganFriendlyi set to true) or false (meaning you can include any restaurant). In addition, you have the filters maxPrice and maxDistance which are the maximum value for price and distance of restaurants you should consider respectively.

Return the array of restaurant IDs after filtering, ordered by rating from highest to lowest. For restaurants with the same rating, order them by id from highest to lowest. For simplicity veganFriendlyi and veganFriendly take value 1 when it is true, and 0 when it is false.

Example 1:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 1, maxPrice = 50, maxDistance = 10
Output: [3,1,5] 
Explanation: 
The restaurants are:
Restaurant 1 [id=1, rating=4, veganFriendly=1, price=40, distance=10]
Restaurant 2 [id=2, rating=8, veganFriendly=0, price=50, distance=5]
Restaurant 3 [id=3, rating=8, veganFriendly=1, price=30, distance=4]
Restaurant 4 [id=4, rating=10, veganFriendly=0, price=10, distance=3]
Restaurant 5 [id=5, rating=1, veganFriendly=1, price=15, distance=1] 
After filter restaurants with veganFriendly = 1, maxPrice = 50 and maxDistance = 10 we have restaurant 3, restaurant 1 and restaurant 5 (ordered by rating from highest to lowest). 

Example 2:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 50, maxDistance = 10
Output: [4,3,2,1,5]
Explanation: The restaurants are the same as in example 1, but in this case the filter veganFriendly = 0, therefore all restaurants are considered.

Example 3:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 30, maxDistance = 3
Output: [4,5]

Constraints:

  • 1 <= restaurants.length <= 10^4
  • restaurants[i].length == 5
  • 1 <= idi, ratingi, pricei, distance<= 10^5
  • 1 <= maxPrice, maxDistance <= 10^5
  • veganFriendlyi and veganFriendly are 0 or 1.
  • All idi are distinct.

Solution

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

C++

花花酱 LeetCode 1332. Remove Palindromic Subsequences

Given a string s consisting only of letters 'a' and 'b'. In a single step you can remove one palindromic subsequence from s.

Return the minimum number of steps to make the given string empty.

A string is a subsequence of a given string, if it is generated by deleting some characters of a given string without changing its order.

A string is called palindrome if is one that reads the same backward as well as forward.

Example 1:

Input: s = "ababa"
Output: 1
Explanation: String is already palindrome

Example 2:

Input: s = "abb"
Output: 2
Explanation: "abb" -> "bb" -> "". 
Remove palindromic subsequence "a" then "bb".

Example 3:

Input: s = "baabb"
Output: 2
Explanation: "baabb" -> "b" -> "". 
Remove palindromic subsequence "baab" then "b".

Example 4:

Input: s = ""
Output: 0

Constraints:

  • 0 <= s.length <= 1000
  • s only consists of letters ‘a’ and ‘b’

Solution: Math

if s is empty => 0 step
if s is a palindrome => 1 step
Otherwise, 2 steps…
1. delete all the as
2. delete all the bs

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

C++

花花酱 LeetCode 1331. Rank Transform of an Array

Given an array of integers arr, replace each element with its rank.

The rank represents how large the element is. The rank has the following rules:

  • Rank is an integer starting from 1.
  • The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
  • Rank should be as small as possible.

Example 1:

Input: arr = [40,10,20,30]
Output: [4,1,2,3]
Explanation: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.

Example 2:

Input: arr = [100,100,100]
Output: [1,1,1]
Explanation: Same elements share the same rank.

Example 3:

Input: arr = [37,12,28,9,100,56,80,5,12]
Output: [5,3,4,2,8,6,7,1,3]

Constraints:

  • 0 <= arr.length <= 105
  • -109 <= arr[i] <= 109

Solution: Sorting + HashTable

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

C++