1011. Capacity To Ship Packages Within D Days¶
A conveyor belt has packages that must be shipped from one port to another within D 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 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
Constraints:
1 <= D <= weights.length <= 5 * 1041 <= weights[i] <= 500
Analysis¶
Using brute-force we can solve it with C(n, D) \approx O(n!), we essentially try all the slots between each two days, and see if we can put all the previous items (up to the last slot that we have picked) in a single conveyor. However, it will cause TLE.
A better approach is not checking all the slots, but check all the possible capacity from max(weights) up to total(weights). Then we can find the minimal capacity that will fullfill our requirement. By using binary search on the capacity, we can speed up the search. Note that this solution is optimal only when total weight is comparable in size of weight array. If max weight and the total weight are not very close and the total weight is much larger than max weight, the search space might be much larger than the burte-force solution. Also keep in mind instead of using 0 as the lower bound, we use max weight. It's just purely for the convenience of the possible() helper function.
Time: O(\log(\sum(weight)) \times n) , where n is the length of the weights array.
Space: O(1)
Code¶
class Solution {
public:
bool inline possible(int cap, vector<int>& w, int D) {
int curr = 0, cnt = 1;
for (int i : w) {
curr += i; // we never check the condition that if i > cap, which will return false
if (curr > cap) {
cnt ++;
curr = i;
}
}
return cnt <= D;
}
int shipWithinDays(vector<int>& w, int D) {
int l = *max_element(w.begin(), w.end()), r = accumulate(w.begin(), w.end(), 0);
while (l < r) {
int cap = l + 0ll + r >> 1;
if (possible(cap, w, D)) {
r = cap;
} else {
l = cap + 1;
}
}
return l;
}
};