Capacity To Ship Packages Within D Days LeetCode

 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

This problem is similar to book alloaction and painter partition problem:

class Solution {

    public int shipWithinDays(int[] weights, int days) {

        // Initialize binary search boundaries

        int low = getMax(weights);    // Min possible capacity (at least max single weight)

        int high = getSum(weights);   // Max possible capacity (sum of all weights)

        int result = 0;               // To store the minimum valid capacity

        // Binary search for the minimum valid capacity

        while (low <= high) {

            int mid = low + (high - low) / 2;  // Midpoint (current capacity guess)

            int dayCount = 1;   // Start with 1 day

            int currentWeight = 0;

            for (int weight : weights) {

                // If adding weight doesn't exceed capacity, keep adding

                if (currentWeight + weight <= mid) {

                    currentWeight += weight;

                } else {

                    // Else, assign to next day

                    dayCount++;

                    currentWeight = weight;

                }

            }

            // If we can ship within given days, try to reduce capacity

            if (dayCount <= days) {

                result = mid;         // Save this as possible result

                high = mid - 1;       // Try smaller capacity

            } else {

                low = mid + 1;        // Need more capacity, shift right

            }

        }

        return result;

    }

    // Helper: get max weight (for lower bound of capacity)

    private int getMax(int[] arr) {

        int max = Integer.MIN_VALUE;

        for (int weight : arr) {

            max = Math.max(weight, max);

        }

        return max;

    }

    // Helper: get total sum of weights (for upper bound of capacity)

    private int getSum(int[] arr) {

        int sum = 0;

        for (int weight : arr) {

            sum += weight;

        }

        return sum;

    }

}