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1011.capacity-to-ship-packages-within-d-days.cpp
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/*
* @lc app=leetcode id=1011 lang=cpp
*
* [1011] Capacity To Ship Packages Within D Days
*
* https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/description/
*
* algorithms
* Medium (53.58%)
* Total Accepted: 11.6K
* Total Submissions: 21.7K
* Testcase Example: '[1,2,3,4,5,6,7,8,9,10]\n5'
*
* 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 <= D <= weights.length <= 50000
* 1 <= weights[i] <= 500
*
*/
class Solution {
public:
bool callme(int ans, vector<int> &wt, int d){
int i = 0;
for(int j=0; j<d; j++){
int sum = 0;
while(i < (int)wt.size() && sum + wt[i] <= ans){
sum += wt[i++];
}
if(i == (int)wt.size()) return true;
}
return i == (int)wt.size();
}
int shipWithinDays(vector<int>& wt, int d) {
int maxi = wt[0];
int sum = 0;
for(int i=0; i<(int)wt.size(); i++)
sum += wt[i], maxi = max(maxi, wt[i]);
int avg = ceil((float)sum/d);
int ans = max(avg, maxi)-1;
for(int k = (ans+sum)/2; k>=1; k/=2){
// cout<<k<<" ";
while(ans+k < sum && !callme(ans+k, wt, d))
ans+=k;
}
return ans+1;
}
};