Using the lightest carriage to move n objects in specific days - algorithm

Question

I have to solve this problem but I can't think of any algorithm for it . Any help is appreciated :)
Problem : we want to move n objects with different weights to another location using a carriage and we have limited days to do so . we can only use the carriage once a day and because it costs a lot we want to choose a carriage that can move all of our product in the given days but pay the minimum amount for it so the objective is to choose a carriage with the least capacity that will do the job ( the price we have to pay for the carriage increases greatly as the capacity goes higher ) . Also the items should be moved according to their weight and the smallest items go first .
the given data : n items , i'th object wights Wi , d days
wanted : c which shows the minimum capacity for our chosen carriage
Example :
10 items
weights : 1 2 3 4 5 6 7 8 9 10
5 days
answer : 15
day 1 -> 1,2,3,4
day 2 -> 5,6
day 3 -> 7,8
day 4 -> 9
day 5 -> 10

Answer 1

To solve this problem efficiently, you should do binary search. You can do binary search on carriage and check for which minimum value you can do that. You can see my C++ solution below for better understanding.
#include<bits/stdc++.h>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef pair<int,int> pii;
//typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
#define fread freopen("input.txt","r",stdin)
#define fwrite freopen("output.txt","w",stdout)
#define eb emplace_back
#define em emplace
#define pb push_back
#define Mp make_pair
#define ff first
#define ss second
#define all(a) a.begin(),a.end()
#define Unique(a) sort(all(a)),a.erase(unique(all(a)),a.end())
#define FastRead ios_base::sync_with_stdio(0);cin.tie(0);
#define memo(ara,val) memset(ara,val,sizeof(ara))
#define II ( { int a ; read(a) ; a; } )
#define LL ( { ll a ; read(a) ; a; } )
#define DD ({double a; scanf("%lf", &a); a;})
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rrep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define pf(n) printf("%lld",n)
#define pfi(n) printf("%d",n)
#define sp printf(" ")
#define ln printf("\n")
#define sc(x) scanf("%lld",&x)
#define scw(x) scanf("%I64d",&x)
#define sci(x) scanf("%d",&x)
#define sc2(x,y) scanf("%lld %lld",&x,&y)
#define sc3(x,y,z) scanf("%lld %lld %lld",&x,&y,&z)
#define Found(a, b) a.find(b) != a.end()
// bool operator< (const node& sx) const { return sx.val < val; }
//set<ll,greater<ll> >st;
//priority_queue<ll , vector<ll> , greater<ll> >
template<class T>inline bool read(T &x){ int c=getchar();int sgn=1;
while(~c&&c<'0'|c>'9'){if(c=='-')sgn=-1;c=getchar();}
for(x=0; ~c&&'0'<=c&&c<='9'; c=getchar())x=x*10+c-'0';x*=sgn;return ~c;
}
const ll N = 200005;
const ll MOD = 1e9+7;
ll ara[N], d, n;
bool check(ll carriageCapacity) {
ll days = 1;
ll s = 0;
bool f = true;
rep1(i, n) {
s += ara[i];
if(s > carriageCapacity) {
days += 1;
s = ara[i];
}
}
if(days > d) f = false;
return f;
}
int main(){
//fread;
//fwrite;
n = LL;
ll sum = 0;
rep1(i,n) {
ara[i] = LL;
sum += ara[i];
}
d = LL;
ll lo = ara[n], hi = sum, ans;
while(lo <= hi) {
ll mid = (lo + hi) >> 1;
if(check(mid)) {
hi = mid - 1;
ans = mid;
}
else {
lo = mid + 1;
}
}
cout << ans << endl;
}

Answer 2

Python solution would look like this.
Using binary search to find the container size that will satisfy the given condition. Complexity of this solution would be O(n * log n)
def shipWithinDays(weights: List[int], D: int) -> int:
def feasible(capacity) -> bool:
days = 1
total = 0
for weight in weights:
total += weight
if total > capacity: # too heavy, wait for the next day
total = weight
days += 1
if days > D: # cannot ship within D days
return False
return True
left, right = max(weights), sum(weights)
while left < right:
mid = left + (right - left) // 2
if feasible(mid):
right = mid
else:
left = mid + 1
return left