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 * 1041 <= weights[i] <= 500Capacity To Ship Packages Within D Days - LeetCode
/*Explanation
Given the number of bags,
return the minimum capacity of each bag,
so that we can put items one by one into all bags.
We binary search the final result.
The left bound is max(A),
The right bound is sum(A).*/
class Solution {
public:
int shipWithinDays(vector<int>& weights, int days) {
int low = *max_element(weights.begin(),weights.end());
int high;
int sum = 0;
for(int w:weights)
sum+=w;
high = sum;
while(low<high)
{
int mid = (low+high)/2;// guess the capacity of ship
int curr_cap = 0;// loaded capacity of current ship
int ships = 1;// number of ship needed
//----simulating loading the weight to ship one by one----
for(int w:weights)
{
curr_cap+=w;
if(curr_cap>mid)// current ship meets its capacity
{
curr_cap = w;
ships++;
}
}
//---------------simulation ends--------------------------
// we need too many ships, so we need to increase capacity to reduce num of ships needed
if(ships>days)
low = mid+1;
// we are able to ship with good num of ships, but we still need to find the optimal max capacity
else
high = mid;
}
return low;
}
};