Check if it is possible to survive on Island

You are a poor person in an island. There is only one shop in this island, this shop is open on all days of the week except for Sunday. Consider following constraints:

  • N – Maximum unit of food you can buy each day.
  • S – Number of days you are required to survive.
  • M – Unit of food required each day to survive.

Currently, it’s Monday, and you need to survive for the next S days.
Find the minimum number of days on which you need to buy food from the shop so that you can survive the next S days, or determine that it isn’t possible to survive.

Examples:



Input : S = 10 N = 16 M = 2
Output : Yes 2
Explanation 1: One possible solution is to buy a box on the first day (Monday), it’s sufficient to eat from this box up to 8th day (Monday) inclusive. Now, on the 9th day (Tuesday), you buy another box and use the chocolates in it to survive the 9th and 10th day.

Input : 10 20 30
Output : No
Explanation 2: You can’t survive even if you buy food because the maximum number of units you can buy in one day is less the required food for one day.

Approach:
In this problem, the greedy approach of buying the food for some consecutive early days is the right direction.
If we can survive for the first 7 days then we can survive any number of days for that we need to check two things
-> Check whether we can survive one day or not.
-> (S >= 7) If we buy food in the first 6 days of the week and we can survive for the week i.e. total food we can buy in a week (6*N) is greater then or equal to total food we require to survive in a week (7*M) then we can survive.
If any of the above conditions are not true then we can’t survive else the minimum number of days required to buy food will be = ceil(total food required/units of food we can buy each day)

CPP

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// C++ program to find the minimum days on which
// you need to buy food from the shop so that you
// can survive the next S days
#include <bits/stdc++.h>
using namespace std;
  
// function to find the minimum days
void survival(int S, int N, int M)
{
  
    // If we can not buy at least a week
    // supply of food during the first week
    // OR We can not buy a day supply of food
    // on the first day then we can't survive.
    if (((N * 6) < (M * 7) && S > 6) || M > N)
        cout << "No\n";
    else {
        // If we can survive then we can
        // buy ceil(A/N) times where A is
        // total units of food required.
        int days = (M * S) / N;
        if (((M * S) % N) != 0)
            days++;
        cout << "Yes " << days << endl;
    }
}
  
// Driver code
int main()
{
    int S = 10, N = 16, M = 2;
    survival(S, N, M);
    return 0;
}

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Java

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// Java program to find the minimum days on which
// you need to buy food from the shop so that you
// can survive the next S days
import java.io.*;
  
class GFG {
  
    // function to find the minimum days
    static void survival(int S, int N, int M)
    {
  
        // If we can not buy at least a week
        // supply of food during the first
        // week OR We can not buy a day supply
        // of food on the first day then we
        // can't survive.
        if (((N * 6) < (M * 7) && S > 6) || M > N)
            System.out.println("No");
  
        else {
  
            // If we can survive then we can
            // buy ceil(A/N) times where A is
            // total units of food required.
            int days = (M * S) / N;
  
            if (((M * S) % N) != 0)
                days++;
  
            System.out.println("Yes " + days);
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int S = 10, N = 16, M = 2;
  
        survival(S, N, M);
    }
}
  
// This code is contributed by vt_m.

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Python3

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# Python3 program to find the minimum days on  
# which you need to buy food from the shop so 
# that you can survive the next S days
def survival(S, N, M):
  
# If we can not buy at least a week 
# supply of food during the first week
# OR We can not buy a day supply of food 
# on the first day then we can't survive.
    if (((N * 6) < (M * 7) and S > 6) or M > N): 
        print("No")
    else:
          
    # If we can survive then we can
    # buy ceil(A / N) times where A is
    # total units of food required.
        days = (M * S) / N
          
        if (((M * S) % N) != 0):
            days += 1
        print("Yes "),
        print(days)
  
# Driver code
S = 10; N = 16; M = 2
survival(S, N, M)
  
# This code is contributed by upendra bartwal

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C#

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// C# program to find the minimum days
// on which you need to buy food from 
// the shop so that you can survive 
// the next S days
using System;
  
class GFG {
  
    // function to find the minimum days
    static void survival(int S, int N, int M)
    {
  
        // If we can not buy at least a week
        // supply of food during the first 
        // week OR We can not buy a day 
        // supply of food on the first day 
        // then we can't survive.
        if (((N * 6) < (M * 7) && S > 6) || M > N)
            Console.Write("No");
        else {
              
            // If we can survive then we can
            // buy ceil(A/N) times where A is
            // total units of food required.
            int days = (M * S) / N;
              
            if (((M * S) % N) != 0)
                days++;
                  
            Console.WriteLine("Yes " + days);
        }
    }
  
    // Driver code
    public static void Main()
    {
        int S = 10, N = 16, M = 2;
          
        survival(S, N, M);
    }
}
  
// This code is contributed by
// Smitha Dinesh Semwal

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PHP

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<?php
// PHP program to find the
// minimum days on which
// you need to buy food 
// from the shop so that you
// can survive the next S days
  
// Function to find 
// the minimum $days
function survival($S, $N, $M)
{
  
    // If we can not buy at least a week
    // supply of food during the first week
    // OR We can not buy a day supply of food
    // on the first day then we can't survive.
    if ((($N * 6) < ($M * 7) && 
          $S > 6) || $M >$N)
        echo "No";
    else 
    {
          
        // If we can survive then we can
        // buy ceil(A/N) times where A is
        // total units of food required.
        $days = ($M * $S) / $N;
        if ((($M * $S) % $N) != 0)
            $days++;
        echo "Yes " , floor($days) ;
    }
}
  
    // Driver code
    $S = 10; $N = 16; $M = 2;
    survival($S, $N, $M);
      
// This code is contributed by anuj_67
  
?>

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Output:

Yes 2

Time Complexity: O(1)
Space Complexity: O(1)



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