1011-capacity-to-ship-packages-within-d-days¶

Description¶

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

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], days = 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], days = 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], days = 4
Output: 3
Explanation:
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Constraints:

1 <= days <= weights.length <= 5 * 10⁴
1 <= weights[i] <= 500

Solution(Python)¶

class Solution:
    def shipWithinDays(self, weights: List[int], days: int) -> int:
        return self.binarySearch(weights, days)

    #
    # we try to fill the ship as maximum as possible
    def feasible(self, capacity: int, weights: List[int], days: int) -> bool:
        days_spent = 1
        cur_weight = 0

        for weight in weights:
            cur_weight += weight
            if cur_weight > capacity:
                days_spent += 1
                cur_weight = weight
            if days_spent > days:
                return False

        return True

    #
    # what will be the range of the leaset weight capacity of the ship?
    #
    # ship must be capable of loading all weights
    # it implies that minimum cap will be maximum weight of the weights list
    #
    #  since we can split a single weight in to multiple rounds
    # maximum possible weight will be sum of all weights in lists
    # atmost we can take all the weights at once which gives
    # upper limit sum(weights)
    #
    # try all possible values between min and max
    # Time Complexity: O(n*M)

    def bruteforce(self, weights: List[int], days: int) -> int:

        mincap = max(weights)
        maxcap = sum(weights)
        for cur_cap in range(mincap, maxcap + 1, 1):
            if self.feasible(cur_cap, weights, days):
                return cur_cap

    # Time Complexity: O(nlog*M)
    def binarySearch(self, weights: List[int], days: int) -> int:
        left = max(weights)
        right = sum(weights)

        while left < right:
            mid = (left + right) >> 1

            if self.feasible(mid, weights, days):
                right = mid
            else:
                left = mid + 1

        return left