Capacity To Ship Packages Within D Days

https://leetcode.com/contest/weekly-contest-128/problems/capacity-to-ship-packages-within-d-days/

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

给定数组划分成m个，问最小化最大子数组和是多少。和Split Array Largest Sum一样。解这种给定背包数，找能把物品一一全部放进去的背包最小容量的题用二分最快。对每个二分mid时检查mid是否能让数组在D次分完，当需要的次数大于D时说明mid取小了，start + 1.

class Solution {
public:
    int check(vector<int>& weights, int m) {
        int sum = 0, c = 1;
        for (int w : weights) {
            if (sum + w > m) {
                sum = 0;
                ++c;
            } 
            sum += w;
        }
        return c;
    }

    int shipWithinDays(vector<int>& weights, int D) {
        int start = 1, end = 500;
        for (int w : weights) start = max(start, w);
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (D < check(weights, mid)) {
                start = mid + 1;
            } else {
                end = mid;
            }    
        }
        return start;
    }
};