Interval Partitioning Problem Implementation

Interval Partitioning Problem Implementation.py

sort the n requests in order of start time
d = 1
assign request 1 to resource 1
min_pq.insert(1, f(i))

for i = 2 to n:
  j = min_pq.minIndex()
  if s(i) >= min_pq.keyOf(j):
    assign request i to resource j
    min_pq.increaseKey(j, f(i))
  else:
    d += 1
    assign request i to resource d
    min_pq.insert(d, f(i))
return d