Doubt at around 3:10 minute

#1

In the video Bhaiya said that we can take the space as 0 and the time taken by the fastest cook to make those parathas but wouldn’t that be too less like say if we have a cook with rank 1 and we want 10 parathas the the space will be just between 0 and 10.

#2

Hello @suhaib0900 please wait let me see the video and clearify what actually sir was saying.

#3

@suhaib0900 sir is saying that to avoid this we will take the range as from 0 to 1e18(infinite) so that we don’t experience anything like that.
if we have 10 paranthas to cook and only 1 cook then the range will be from 1 to 55 as the single cook will take 55 minutes to cook all the paranthas but here there can be more then one cook so we can eventually reduce this time much more.

#6

Hi Can you please tell me what is wrong with this code for the Roti Parata problem.

#include <stdio.h>
#include
#include
#include
using namespace std;
//c- num of chef
//req- requirement paratha
//rank - rank of chefs
bool solve(int rank[], int req, int c, int mid)
{
int paratha = 0;

for (int i = 0; i < c; i++)
{
	for (int p = rank[i]; p <= mid; p = p + rank[i])
	{
		paratha++;
	}
}
//cout << endl << "P=" << paratha << endl;
if (paratha >= req)
	return true;

return false;

}
int main() {
// your code goes here
#ifndef ONLINE_JUDGE
freopen(“input.txt”, “r”, stdin); // redirects standard input
freopen(“output.txt”, “w”, stdout); // redirects standard output
#endif
int t;
cin >> t;
while (t–)
{

	int req;
	cin >> req;//Kitna Paratha
	int c;
	cin >> c;//Number of cooks
	int rank[c];
	for (int i = 0; i < c; i++)
	{
		cin >> rank[i];
	}
	sort(rank, rank + c);
	int ans = 0;
	int s = 0;
	/*int e = 0;
	int temp = 0; //temporary request for paratha
	while (temp <= req)
	{
		e = e + rank[0];
		cout << "e = " << e << endl;
		temp++;
	}*/
	int e = (req * req) / 2;

	//	cout << endl << "Val = " << e << endl;
	while (s <= e)
	{	int mid = s + (e - s) / 2;
		//cout << mid;
		if (solve(rank, req, c, mid))
		{
			ans = mid;

			e = mid - 1;
		}
		else
		{
			s = mid + 1;
		}
	}
	cout << ans << endl;
}
return 0;

}

#7

@suhaib0900 check this:

#8

Hi Tushar.
I had a doubt. In the code that you shared are we considering that there will always be a chef with rank 1?

#9

Hello @suhaib0900 read this explanation of the ouput i hope this helps you:
First cook with rank 1 cooks 4 paranthas in 10 minutes (1+2+3+4).
Second cook with rank 2 cooks 3 paranthas in 12 minutes (2+4+6)
Third cook with rank 3 cooks 2 paranthas in 9 minutes (3+6) Fourth cook with rank 4 only needs to cook one last remaining parantha. He can do that in 4 minutes.
Since these cooks cook parallely, the total time taken will be the maximum of the four i.e. 12 minutes.

#10

Okay got it. Thanks! You can close the thread.

#11

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