1011.Capacity To Ship Packages Within D Days
===

###### tags: `Medium`,`Array`,`Binary Search`

[1011. Capacity To Ship Packages Within D Days](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)

### 題目描述

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.

### 範例

**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 * 10^4^
* 1 <= `weights[i]` <= 500

### 解答
#### Python
##### 思路 :
每日需運輸的最小重量至少會是陣列的最大值，最大重量則是陣列的總和
對這之間的重量去做 Binary Search，看看能不能在天數內搬運完所有的貨物

class Solution:
    def shipWithinDays(self, weights: List[int], days: int) -> int:
        def feasible(capacity):
            total, shipping_day = 0, 1

            for weight in weights:
                total += weight
                if total > capacity:
                    total = weight
                    shipping_day += 1

            return shipping_day <= days

        l, r = max(weights), sum(weights)
        while l < r:
            mid = l + (r - l) // 2
            if feasible(mid):
                r = mid
            else:
                l = mid + 1

        return l

> [name=Ron Chen][time=Wed, Feb 22, 2022]

class Solution {
public:
    int shipWithinDays(vector<int>& weights, int days) {
        int left = *max_element(begin(weights), end(weights)), right = accumulate(begin(weights), end(weights), 0);
        while(left < right)
        {
            int mid = left + (right-left)/2;
            int cur_days = 1, cur_sum = 0;
            for(int w : weights)
            {
                cur_sum += w;
                if(cur_sum > mid)
                {
                    cur_days++;
                    cur_sum = w;
                }
            }
            if(cur_days > days)
                left = mid+1;
            else
                right = mid;
        }
        return left;
    }
};

Time: $O(n*log(500*n))$ 確認 $n$ 個element最多 $log(運輸重量最大值)$ 次
Extra Space: $O(1)$

> [name=XD] [time= Feb 22, 2023]

### Reference

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