Formatted question description: https://leetcode.ca/all/1011.html

# 1011. Capacity To Ship Packages Within D Days (Medium)

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


Note:

1. 1 <= D <= weights.length <= 50000
2. 1 <= weights[i] <= 500

Companies:

Related Topics:
Array, Binary Search

## Solution 1.

// OJ: https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/
// Time: O(W * L) where W is the sum of all weights, and L is the length of weights.
// Space: O(1)
class Solution {
private:
bool ok(vector<int>& weights, int D, int capacity) {
int d = 1, c = 0;
for (int w : weights) {
if (c + w > capacity) {
c = 0;
++d;
if (d > D) return false;
}
c += w;
}
return true;
}
public:
int shipWithinDays(vector<int>& weights, int D) {
int sum = 0, maxVal = 0;
for (int w : weights) {
sum += w;
maxVal = max(maxVal, w);
}
int capacity = max((sum + D - 1) / D, maxVal);
while (!ok(weights, D, capacity)) capacity++;
return capacity;
}
};


Java

• class Solution {
public int shipWithinDays(int[] weights, int D) {
int sum = 0;
int low = 0;
for (int weight : weights) {
sum += weight;
low = Math.max(low, weight);
}
int high = sum;
while (low < high) {
int mid = (high - low) / 2 + low;
int days = daysNeed(weights, mid);
if (days > D)
low = mid + 1;
else
high = mid;
}
return low;
}

public int daysNeed(int[] weights, int capacity) {
int days = 1;
int currentTotal = 0;
int length = weights.length;
for (int i = 0; i < length; i++) {
int weight = weights[i];
if (currentTotal + weight <= capacity)
currentTotal += weight;
else {
days++;
currentTotal = weight;
}
}
return days;
}
}

• Todo

• # 1011. Capacity To Ship Packages Within D Days
# https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/

class Solution:
def shipWithinDays(self, weights: List[int], days: int) -> int:
total = sum(weights)

def good(target):
mmax = s = 0
day = 1

for x in weights:
if s + x > target:
s = x
day += 1
else:
s += x
mmax = max(mmax, s)

return day <= days and mmax <= target

left, right = 1, total

while left < right:
mid = left + (right - left) // 2

if good(mid):
right = mid
else:
left = mid + 1

return left