min_diff(i, d)
representing the minimum difficulty starting from job i
with d
days left.d
is 1, return the maximum difficulty of jobs from i
to the end.i
is equal to the length of the job array, return 0 if d
is 0, or -1
otherwise.j
from i
to the end, and for each j
, calculate the difficulty of scheduling jobs from i
to j
on the current day and add it to the result of min_diff(j, d - 1)
.i
with d
days left.