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

Problem Statement

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.

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

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:

C++

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;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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