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.
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Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2020/01/16/untitled.png\")
\n\nInput: jobDifficulty = [6,5,4,3,2,1], d = 2\nOutput: 7\nExplanation: First day you can finish the first 5 jobs, total difficulty = 6.\nSecond day you can finish the last job, total difficulty = 1.\nThe difficulty of the schedule = 6 + 1 = 7 \n\n\n
Example 2:
\n\n\nInput: jobDifficulty = [9,9,9], d = 4\nOutput: -1\nExplanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.\n\n\n
Example 3:
\n\n\nInput: jobDifficulty = [1,1,1], d = 3\nOutput: 3\nExplanation: The schedule is one job per day. total difficulty will be 3.\n\n\n
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Constraints:
\n\n1 <= jobDifficulty.length <= 3000 <= jobDifficulty[i] <= 10001 <= d <= 10Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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