problem1011.py

'''

1011. Capacity To Ship Packages Within D Days
Medium

A conveyor belt has packages that must be shipped from one port to another within D days.

The i-th package on the conveyor belt has a weight of weights[i].  Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

 

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation: 
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10

Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. 

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation: 
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation: 
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Solution:
Use bipart to find the answer which should be in between the largest value and the smallest value.
The largest partition value is the sum of all cargos, the smallest is the max alue of the cargo.
Corresbondingly, the smallest partition number is 1, the biggest partition number is the number of the cargos.

'''
class Solution(object):
    def shipWithinDays(self, weights, D):
        """
        :type weights: List[int]
        :type D: int
        :rtype: int
        """
        def partition(weights, capacity):
            rst = 0
            currentLoad = 0
            for i in range(len(weights)):
                if weights[i] + currentLoad > capacity:
                    rst += 1
                    currentLoad = 0
                currentLoad += weights[i]
            return rst + 1
        if not weights:
            return 0
        leastWeight = max(weights)
        largestWeight = sum(weights)
        if D == 1:
            return largestWeight

        while leastWeight < largestWeight:
            load = (leastWeight + largestWeight) / 2
            if load == leastWeight or load == largestWeight:
                break
            currentD = partition(weights, load)
            if currentD > D:
                #The partition is too many, weight too small
                leastWeight = load
            elif currentD <= D:
                #The partition is too few, weight too big
                largestWeight = load

        if partition(weights, leastWeight) <= D:
            return leastWeight
        return largestWeight

s = Solution()

weights = [1,2,3,4,5,6,7,8,9,10]
D = 10
print s.shipWithinDays(weights, D)