Cake Distribution Problem

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: 0 
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: 1 
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)

Auxiliary Space: O(1)

Approach name: Recursive approach

Steps:

• Define a recursive function “calculate_left_cakes” that takes three parameters “n”, “m”, and “i“.
• If the value of “m” is less than the value of “i”, then return the value of “m” minus 1.
• Otherwise, subtract the value of “i” from “m”, increment the value of “i” by 1, and call the recursive function again with the new values of “m” and “i”.
• Finally, print the result of the function call.

Below is the implementation of above approach:-

C++

#include <cmath>
#include <iostream>
using namespace std;

int cntCakes(int n, int m)
{
    int sum = (n * (n + 1)) / 2;
    int quo = m / sum;
    int rem = m % sum;
    int ans = m - quo * sum;
    int x = (-1 + sqrt(1 + 8 * rem)) / 2;
    ans = ans - x * (x + 1) / 2;
    return ans;
}

int main()
{
    int n = 4;
    int m = 11;
    int ans = cntCakes(n, m);
    cout << ans << endl;
    return 0;
}

Java

import java.util.*;

public class Main {
    public static int cntCakes(int n, int m) {
        int sum = (n * (n + 1)) / 2;
        int quo = m / sum;
        int rem = m % sum;
        int ans = m - quo * sum;
        int x = (int) ((-1 + Math.sqrt(1 + 8 * rem)) / 2);
        ans = ans - x * (x + 1) / 2;
        return ans;
    }

    public static void main(String[] args) {
        int n = 4;
        int m = 11;
        int ans = cntCakes(n, m);
        System.out.println(ans);
    }
}

Python3

def calculate_left_cakes(n, m, i=1):
    if m < i:
        return m-1
    return calculate_left_cakes(n, m-i, i+1)

# example usage
n = 4
m = 11
left_cakes = calculate_left_cakes(n, m)
print("Left Cakes:", left_cakes)

n = 3
m = 8
left_cakes = calculate_left_cakes(n, m)
print("Left Cakes:", left_cakes)

C#

using System;

public class Program {
    public static int cntCakes(int n, int m) {
        int sum = (n * (n + 1)) / 2;
        int quo = m / sum;
        int rem = m % sum;
        int ans = m - quo * sum;
        int x = (int) ((-1 + Math.Sqrt(1 + 8 * rem)) / 2);
        ans = ans - x * (x + 1) / 2;
        return ans;
    }

    public static void Main(string[] args) {
        int n = 4;
        int m = 11;
        int ans = cntCakes(n, m);
        Console.WriteLine(ans);
    }
}

Javascript

function cntCakes(n, m) {
    let sum = (n * (n + 1)) / 2;
    let quo = Math.floor(m / sum);
    let rem = m % sum;
    let ans = m - quo * sum;
    let x = Math.floor((-1 + Math.sqrt(1 + 8 * rem)) / 2);
    ans = ans - x * (x + 1) / 2;
    return ans;
}

let n = 4;
let m = 11;
let ans = cntCakes(n, m);
console.log(ans);

Output

0

Time complexity: O(n), where n is the value of “m”. In the worst case, the recursive function will be called n times before the base case is reached.
Auxiliary Space: O(1)