A laundry store cleans laundry from $n$ customers $C_1,C_2,dots{C}_n$ in a particular day. The

clothes from each customer must be first washed and then dried on the store's machines.

However, for the current day all except one of the washing machines are scheduled for

maintenance, thus there is only one washing machine available for use. After washing,

there are sufficient drying machines to dry the clothes from all customers in parallel, and

they are all operational. For customer $C_i,$ the laundry needs $w_i$ minutes to be washed and $d_i$

minutes to be dried. The store needs to come up with a schedule in which the customers'

clothes are to be washed in the single machine. The cleaning time of the schedule is the

earliest time by when all the clothes have been washed and dried. Give a polynomial$-$

time algorithm that finds a schedule with a minimum cleaning time.

Note: to prove that your greedy strategy yields the optimal solution, you have to prove that

the problem has the greedy$-$choice property.