Monday, February 28, 2022

LeetCode 1335. Minimum Difficulty of a Job Schedule (Iteration in the recurrence relation)

1335. Minimum Difficulty of a Job Schedule

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You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the ith 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 on that day.

You are given an integer array jobDifficulty and an integer d. The difficulty of the ith 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.

 

Constraints:

  • 1 <= jobDifficulty.length <= 300
  • 0 <= jobDifficulty[i] <= 1000
  • 1 <= d <= 10
class Solution {
public:
/**
Framework (Top-Down: Memorization)
1. function
dp[i][day]: min diff, starts ith job, with day
dp[0][1]: answer
2. relation
min(hardest + dp[i+1][j+1]), 1 < j < n - day
3. base
(day == d) max(jobs[i..])
*/
int D, n;
vector<int> jobs;
vector<int> hardestRemain;
vector<vector<int>> memo;
int dp(int i, int day){
// base
if(day == D)
return hardestRemain[i];
if(memo[i][day] == - 1){
int limit = n - (D - day);
int ans = 10001;
int hardest = 0;
for(int j = i; j < limit; j++){
hardest = max(hardest, jobs[j]);
ans = min(ans, dp(j + 1, day + 1) + hardest);
}
memo[i][day] = ans;
}
return memo[i][day];
}
int minDifficulty(vector<int>& jobDifficulty, int d) {
// make global
D = d;
jobs = jobDifficulty;
n = jobs.size();
hardestRemain.resize(n);
memo.resize(n+1, vector<int>(d+1, -1));
// edge case
if(d > n) return -1;
// prepare hardest remain
for(int i = n - 1; i >= 0; i--){
hardestRemain[i] = jobs[i];
if(i < n - 1) hardestRemain[i] = max(hardestRemain[i+1], hardestRemain[i]);
}
return dp(0, 1);
}
};

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