LeetCode 1335. Minimum Difficulty of a Job Schedule Solution in Java, C++, Python & More | Explanation + Code

Description

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

Solutions

Solution 1: Dynamic Programming

We define f[i][j] as the minimum difficulty to finish the first i jobs within j days. Initially f[0][0] = 0, and all other f[i][j] are ∞.

For the j-th day, we can choose to finish jobs [k,..i] on this day. Therefore, we have the following state transition equation:

f[i][j] = min_{k ∈ [1,i]} {f[k-1][j-1] + max_{k ≤ t ≤ i} {jobDifficulty[t]}}

The final answer is f[n][d].

The time complexity is O(n² × d), and the space complexity is O(n × d). Here n and d are the number of jobs and the number of days respectively.

PythonJavaC++GoTypeScript

class Solution: def minDifficulty(self, jobDifficulty: List[int], d: int) -> int: n = len(jobDifficulty) f = [[inf] * (d + 1) for _ in range(n + 1)] f[0][0] = 0 for i in range(1, n + 1): for j in range(1, min(d + 1, i + 1)): mx = 0 for k in range(i, 0, -1): mx = max(mx, jobDifficulty[k - 1]) f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx) return -1 if f[n][d] >= inf else f[n][d](code-box)

class Solution { public int minDifficulty(int[] jobDifficulty, int d) { final int inf = 1 << 30; int n = jobDifficulty.length; int[][] f = new int[n + 1][d + 1]; for (var g : f) { Arrays.fill(g, inf); } f[0][0] = 0; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= Math.min(d, i); ++j) { int mx = 0; for (int k = i; k > 0; --k) { mx = Math.max(mx, jobDifficulty[k - 1]); f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx); } } } return f[n][d] >= inf ? -1 : f[n][d]; } }(code-box)

class Solution { public: int minDifficulty(vector<int>& jobDifficulty, int d) { int n = jobDifficulty.size(); int f[n + 1][d + 1]; memset(f, 0x3f, sizeof(f)); f[0][0] = 0; for (int i = 1; i <= n; ++i) { for (int j = 1; j <= min(d, i); ++j) { int mx = 0; for (int k = i; k; --k) { mx = max(mx, jobDifficulty[k - 1]); f[i][j] = min(f[i][j], f[k - 1][j - 1] + mx); } } } return f[n][d] == 0x3f3f3f3f ? -1 : f[n][d]; } };(code-box)

func minDifficulty(jobDifficulty []int, d int) int { n := len(jobDifficulty) f := make([][]int, n+1) const inf = 1 << 30 for i := range f { f[i] = make([]int, d+1) for j := range f[i] { f[i][j] = inf } } f[0][0] = 0 for i := 1; i <= n; i++ { for j := 1; j <= min(d, i); j++ { mx := 0 for k := i; k > 0; k-- { mx = max(mx, jobDifficulty[k-1]) f[i][j] = min(f[i][j], f[k-1][j-1]+mx) } } } if f[n][d] == inf { return -1 } return f[n][d] }(code-box)

function minDifficulty(jobDifficulty: number[], d: number): number { const n = jobDifficulty.length; const inf = 1 << 30; const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(d + 1).fill(inf)); f[0][0] = 0; for (let i = 1; i <= n; ++i) { for (let j = 1; j <= Math.min(d, i); ++j) { let mx = 0; for (let k = i; k > 0; --k) { mx = Math.max(mx, jobDifficulty[k - 1]); f[i][j] = Math.min(f[i][j], f[k - 1][j - 1] + mx); } } } return f[n][d] < inf ? f[n][d] : -1; }(code-box)