Dividing B cheese cakes among N classes to minimize maximum no of students per cake - algorithm

Question

There is a school in a village. It has N classes. One fine day, someone donated B blue berry cheese cakes to schools. Now you need to divide these cakes such that:
Each class gets at least 1 cake.
Each class will share the cake(s) among students.
Your aim is to minimize the maximum number of students per cake in any class.
input
contains two space separated integers N and B denoting the number of classes and total number of blue berry cheese cakes, respectively.
Next N lines contain number of students in each class.
Output
For each test case, output the maximum number of students who will share a cake.
Constraints
2 <= N <= 5*10^5
N <= B <= 2*10^6 1 <= number of students in ith class <= 5*10^6
Sample Input - 1 1 2 35 Sample Output - 1 18 Sample Input - 2 2 7 20 50 Sample Output - 2 10

Answer 1

Apply binary search to find minimum number of students/cake.
Take lower limit 'l' as 1 and upper limit 'u' as max number of students in any given class.
Let 'mid' represents current students/cake where mid=(l+u)/2, then,
calculate required number of cakes 'req' for each mid value,
if req<=total number of cakes u=mid-1 else l=mid+1.
Apply the binary search.
Handle the case when n>b separately.
Link to my C++ code for this problem : https://ideone.com/zsTHC5
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
ll n,b;cin>>n>>b;
ll arr[n],ans=0;
for(ll i=0;i<n;i++){cin>>arr[i];ans=max(ans,arr[i]);}
if(n>b)cout<<"-1";
else if(n==b)cout<<ans;
else // binary search to find minimum students/cake
{
ll l=1,r=ans,mid; // mid represents current number of students/cake
while(l<=r)
{
mid=(l+r)/2;
ll ct=0;
for(ll i=0;i<n;i++)
{
ct+=(arr[i]+mid-1)/mid;
}
if(ct<=b)
{
ans=min(ans,mid);
r=mid-1;
}
else
l=mid+1;
}
cout<<ans;
}
return 0;
}

Answer 2

I agree with smartsn123.The solution is pretty simple ,initialize the max heap with each class with one cake and then start distributing cakes one by one.Here following the java implementation of the solution.
import java.util.*;
class Node{
private int nStu ;
private int nCake;
public Node(int nStu , int nCake ){
this.nStu = nStu;
this.nCake = nCake;
}
public int getnStu(){
return nStu;
}
public int getnCake(){
return nCake;
}
public void addCake(){
nCake++;
}
}
class CheeseCake{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int B = sc.nextInt();
int[] arr = new int[N];
int ext = B - N ;
MyComparator mc = new MyComparator();
PriorityQueue<Node> queue = new PriorityQueue<>(mc);
for(int i = 0 ; i < N ; i++){
//arr[i] = sc.nextInt();
int temp = sc.nextInt();
Node newNode = new Node(temp , 1);
queue.add(newNode);
}
while(ext != 0){
Node new1 = queue.poll();
new1.addCake();
queue.add(new1);
ext--;
}
Node newNode1 = queue.poll();
int fStu = newNode1.getnStu();
int fCake = newNode1.getnCake();
System.out.println((fStu+1)/2);
}
}
class MyComparator implements Comparator<Node>{
#Override
public int compare(Node n1 , Node n2){
/*arranging based on the students per cakes*/
double avg1 = n1.getnStu() / n1.getnCake();
double avg2 = n2.getnStu() / n2.getnCake();
if(avg1 > avg2){
return -1;
}
if(avg1 < avg2){
return 1;
}
return 0;
}
}