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1011. Capacity To Ship Packages Within D Days 
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 * 1041 <= weights[i] <= 500
Click to open Hints
- Binary search on the answer. We need a function possible(capacity) which returns true if and only if we can do the task in D days.
Solution:
rs
impl Solution {
pub fn ship_within_days(weights: Vec<i32>, days: i32) -> i32 {
let mut min_cap = 0; // it is the max of all weights
let mut max_cap = 0; // sum of all weights
for w in weights.iter() {
// assign the max of all weights to min_cap
min_cap = min_cap.max(*w);
// add all the weights to right
max_cap += w;
}
// binary search
while min_cap < max_cap {
let mid = (min_cap + max_cap) / 2;
let mut days_used = 1;
let mut cur_cap = 0;
// for each weight, if the current capacity + weight is greater than mid,
// then we need to use another day
// otherwise, we can use the current day
for w in weights.iter() {
if cur_cap + w > mid {
days_used += 1;
cur_cap = 0;
}
cur_cap += w;
}
// if days_used is greater than days, then we need to increase the capacity
if days_used > days {
min_cap = mid + 1;
} else {
max_cap = mid;
}
}
min_cap
}
}
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