Formatted question description: https://leetcode.ca/all/1335.html

1335. Minimum Difficulty of a Job Schedule (Hard)

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

Related Topics:
Dynamic Programming

Solution 1. DP

Let dp[d][i] be the answer for the subproblem with d days at ith job.

Let mx[i][j] be the maximum value in A[i..j].

dp[d][i] = min( dp[d-1][j-1] + mx[j][i] | d-1 <= j <= i )
// OJ: https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/

// Time: O(N^2 * D)
// Space: O(N^2 + ND)
class Solution {
    typedef long long LL;
    inline void setMin(LL &a, LL b) { a = min(a, b); }
public:
    int minDifficulty(vector<int>& A, int D) {
        int N = A.size();
        if (D > N) return -1;
        vector<vector<LL>> mx(N, vector<LL>(N)), dp(D + 1, vector<LL>(N, 1e9));
        for (int i = 0; i < N; ++i) {
            for (int j = i; j < N; ++j) mx[i][j] = *max_element(A.begin() + i, A.begin() + j + 1);
        }
        for (int i = 0; i < N; ++i) dp[1][i] = mx[0][i];
        for (int d = 2; d <= D; ++d) {
            for (int i = d - 1; i < N; ++i) {
                for (int j = d - 1; j <= i; ++j) {
                    setMin(dp[d][i], dp[d - 1][j - 1] + mx[j][i]);
                }
            }
        }
        return dp[D][N - 1];
    }
};

Solution 2. DP

We can compute the mx while computing dp instead of computing mx array beforehand.

// OJ: https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/

// Time: O(N^2 * D)
// Space: O(ND) 
class Solution {
    typedef long long LL;
    inline void setMin(LL &a, LL b) { a = min(a, b); }
public:
    int minDifficulty(vector<int>& A, int D) {
        int N = A.size(), inf = 1e9;
        if (D > N) return -1;
        vector<vector<LL>> dp(D + 1, vector<LL>(N, inf));
        for (int i = 0; i < N; ++i) dp[1][i] = i == 0 ? A[0] : max(dp[1][i - 1], (LL)A[i]);
        for (int d = 2; d <= D; ++d) {
            for (int i = d - 1; i < N; ++i) {
                int mx = 0;
                for (int j = i; j >= d - 1; --j) {
                    mx = max(mx, A[j]);
                    setMin(dp[d][i], dp[d - 1][j - 1] + mx);
                }
            }
        }
        return dp[D][N - 1];
    }
};

Solution 3. DP

Since dp[d][i] is dependent on dp[d-1][j-1] and j <= i, we can flip the loop direction and just need 1D dp array.

// OJ: https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/

// Time: O(NND)
// Space: O(N)
class Solution {
    typedef long long LL;
    inline void setMin(LL &a, LL b) { a = min(a, b); }
public:
    int minDifficulty(vector<int>& A, int D) {
        int N = A.size(), inf = 1e9;
        if (D > N) return -1;
        vector<LL> dp(N);
        for (int i = 0; i < N; ++i) dp[i] = i == 0 ? A[0] : max(dp[i - 1], (LL)A[i]);
        for (int d = 2; d <= D; ++d) {
            for (int i = N - 1; i >= d - 1; --i) {
                int mx = 0;
                dp[i] = inf;
                for (int j = i; j >= d - 1; --j) {
                    mx = max(mx, A[j]);
                    setMin(dp[i], dp[j - 1] + mx);
                }
            }
        }
        return dp[N - 1];
    }
};

Solution 4. DP + Monotonic Min Stack

TODO https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/discuss/490316/JavaC%2B%2BPython3-DP-O(nd)-Solution

https://leetcode.com/problems/minimum-difficulty-of-a-job-schedule/discuss/495000/C%2B%2B-0ms!-O(d*n)-time-O(n)-space.-DP-%2B-MonotonicMinimum-Stack

Java

class Solution {
    public int minDifficulty(int[] jobDifficulty, int d) {
        if (jobDifficulty == null)
            return -1;
        int length = jobDifficulty.length;
        if (length < d)
            return -1;
        if (length == d) {
            int sum = 0;
            for (int difficulty : jobDifficulty)
                sum += difficulty;
            return sum;
        }
        int[] maxDifficultyRight = new int[length];
        maxDifficultyRight[length - 1] = jobDifficulty[length - 1];
        for (int i = length - 2; i >= 0; i--)
            maxDifficultyRight[i] = Math.max(maxDifficultyRight[i + 1], jobDifficulty[i]);
        int[][] dp = new int[d][length];
        dp[0][0] = jobDifficulty[0];
        for (int i = 1; i <= length - d; i++)
            dp[0][i] = Math.max(dp[0][i - 1], jobDifficulty[i]);
        for (int i = 1; i < d; i++) {
            int start = i, end = length - d + i;
            int prevEnd = start - 1;
            for (int j = start; j <= end; j++) {
                dp[i][j] = dp[i - 1][j - 1] + jobDifficulty[j];
                int max = jobDifficulty[j];
                for (int k = j - 2; k >= prevEnd; k--) {
                    max = Math.max(max, jobDifficulty[k + 1]);
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][k] + max);
                }
            }
        }
        return dp[d - 1][length - 1];
    }
}

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