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# 1011. Capacity To Ship Packages Within D Days / Book Allocation Problem

## Problem Statement

<br>

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

The `ith` 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.

&#x20;

**Example 1:**

<pre><code><strong>Input: weights = [1,2,3,4,5,6,7,8,9,10], days = 5
</strong><strong>Output: 15
</strong><strong>Explanation: A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
</strong>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.
</code></pre>

**Example 2:**

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

**Example 3:**

<pre><code><strong>Input: weights = [1,2,3,1,1], days = 4
</strong><strong>Output: 3
</strong><strong>Explanation:
</strong>1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1
</code></pre>

&#x20;

**Constraints:**

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

## Intuition

```
Approach:

Here BS from -> low = max element as we have to take
That element as a whole so 

high = addition of all 

low -> high BS
```

### Links

<https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/description/>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int shipWithinDays(vector<int>& arr, int days) {
        int low=*max_element(arr.begin(), arr.end()); 
        int high=accumulate(arr.begin(), arr.end(), 0);
        int ans;

        while(low<=high){
            int mid = low+(high-low)/2;
            int temp_days=0;
            int temp_sum=0;
            for(auto &it: arr){
                temp_sum += it;
                if(temp_sum > mid){
                    temp_days++;
                    temp_sum = it;
                } 
            }

            temp_days++;

            if(temp_days <= days){
                ans = mid;
                high = mid-1;
            }
            else
                low = mid+1;
        }

        return ans;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###