Skip to content
Related Articles

Related Articles

Improve Article
Cake Distribution Problem
  • Difficulty Level : Easy
  • Last Updated : 31 May, 2021

Given two integers N and M, where N is the number of friends sitting in a clockwise manner in a circle and M is the number of cakes. The task is to calculate the left number of cakes after distributing i cakes to i’th friend. The distribution of cakes will stop if the count of cakes is less than the required amount.
Examples: 
 

Input: N = 4, M = 11 
Output:
1st round: 
The 1st friend gets 1 cake, 2nd gets 2 cakes, 
3rd get 3 and 4th gets 4 cakes. 
Remaining cakes = 11 – (1 + 2 + 3 + 4) = 1 
2nd round: 
This time only 1st friend gets the left 1 cake. 
Remaining cakes = 1 – 1 = 0
Input: N = 3, M = 8 
Output:
1st round: 
The 1st friend gets 1 cake, 2nd gets 2 cakes, 
and 3rd get 3 cakes. 
Remaining cakes = 8 – (1 + 2 + 3) = 2 
2nd round: 
This time only 1st friend gets the left 1 cake, 
and then there is no cake left for 2nd friend. 
Remaining cakes = 2 – 1 = 1 
 

 

Approach: 
 

  • Check how many cycles of distribution of cakes are possible from m number of cakes.
  • Calculate the number of cakes for 1 cycle which is 
     
sum = n * (n + 1) / 2
  • Now diving M by sum we get cycle count + some remainder.
  • Now check how many remaining cakes are again possible to distribute to x friends.
  • The value of x can be easily achieved by solving quadratic equation 
     
remainder = x * (x + 1) / 2

Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to return the
// remaining count of cakes
int cntCakes(int n, int m)
{
 
    // Sum for 1 cycle
    int sum = (n * (n + 1)) / 2;
 
    // no. of full cycle and remainder
    int quo = m/sum ;
    int rem = m % sum ;
    double ans = m - quo * sum ;
 
    double x = (-1 + pow((8 * rem) + 1, 0.5)) / 2;
     
    ans = ans - x * (x + 1) / 2;
 
    return int(ans);
}
 
// Driver Code
int main ()
{
    int n = 3;
    int m = 8;
    int ans = cntCakes(n, m);
    cout << (ans);
}
 
// This code is contributed by Surendra_Gangwar

Java




// Java implementation of the approach
class GFG
{
     
    // Function to return the
    // remaining count of cakes
    static int cntCakes(int n, int m)
    {
     
        // Sum for 1 cycle
        int sum = (n * (n + 1)) / 2;
     
        // no. of full cycle and remainder
        int quo = m/sum ;
        int rem = m % sum ;
        double ans = m - quo * sum ;
     
        double x = (-1 + Math.pow((8 * rem) + 1, 0.5)) / 2;
         
        ans = ans - x * (x + 1) / 2;
     
        return (int)ans;
    }
 
    // Driver Code
    public static void main (String[] args)
    {
        int n = 3;
        int m = 8;
        int ans = cntCakes(n, m);
        System.out.println(ans);
    }
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach
 
# Function to return the
# remaining count of cakes
def cntCakes(n, m):
 
    # Sum for 1 cycle
    sum = (n*(n + 1))//2
 
    # no. of full cycle and remainder
    quo, rem = m//sum, m % sum
    ans = m - quo * sum
 
    x = int((-1 + (8 * rem + 1)**0.5)/2)
    ans = ans - x*(x + 1)//2
 
    return ans
 
# Driver code
def main():
  n = 4
  m = 11
  ans = cntCakes(n, m)
  print(ans)
 
main()

C#




// C# implementation of the approach
using System;
 
class GFG
{
    // Function to return the
    // remaining count of cakes
    static int cntCakes(int n, int m)
    {
     
        // Sum for 1 cycle
        int sum = (n * (n + 1)) / 2;
     
        // no. of full cycle and remainder
        int quo = m/sum ;
        int rem = m % sum ;
        double ans = m - quo * sum ;
     
        double x = (-1 + Math.Pow((8 * rem) + 1, 0.5)) / 2;
         
        ans = ans - x * (x + 1) / 2;
     
        return (int)ans;
    }
 
    // Driver Code
    static public void Main ()
    {
        int n = 3;
        int m = 8;
        int ans = cntCakes(n, m);
        Console.Write(ans);
    }
}
 
// This code is contributed by ajit.

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the
    // remaining count of cakes
    function cntCakes(n, m)
    {
       
        // Sum for 1 cycle
        let sum = (n * (n + 1)) / 2;
       
        // no. of full cycle and remainder
        let quo = m/sum;
        let rem = m % sum ;
        let ans = m - quo * sum + 6;
       
        let x = (-1 + Math.pow((8 * rem) + 1, 0.5));
           
        ans = ans - x * (x + 1) / 2;
       
        return parseInt(ans, 10);
    }
     
    let n = 3;
    let m = 8;
    let ans = cntCakes(n, m);
    document.write(ans);
 
// This code is contributed by suresh07.
</script>
Output: 
0

 

Time Complexity: O(1)
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :