Calculate MDAS Factorial of given number

Given an integer N, the task is to find the MDAS factorial.

The general factorial of a no. N is given by :

Factorial(N) = (N)*(N-1)*(N-2)*(N-3)*(N-4)*(N-5)*(N-6)*(N-7)- – – – – -(3)*(2)*(1).



In MDAS factorial, instead of simply multiplying the numbers from N to 1, we perform four operations, Multiplication(*), Divide(/), Addition(+) and Subtraction(-) in a repeating pattern as shown below:

MDAS_Factorial(N) = (N) * (N-1) / (N-2) + (N-3) – (N-4) – – – – – upto 1.

By using the integers in decreasing order, we swap the multiplication operations for fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in the above order.

Examples:

Input : N = 4
Output : 7
Explaination : MDAS_Factorial(4) = 4 * 3 / 2 + 1 = 7

Input : N = 10
Output : 12
Explaination : 
MDAS_Factorial(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1 = 12

Simple Approach: : The idea is to use a loop for each cycle of operations (*,/,+,-) and calculate the MDAS_Factorial of N. But this may work slow if N is very large. Time Complexity of this approach is O(N).

Efficient Approach :
If we observe carefully it can be concluded that:

  1. If N is less than or equal to 2 then answer will be N itself.
  2. If N is 3 OR N is 4, the answer is N + 3.
  3. If (N – 4) is completely divisible by 4, the answer is N + 1.
  4. If (N – 4) gives remainder 1 OR 2 while dividing by 4, the answer is N + 2.
  5. For the remaining values, the answer will be N – 1.

Below is the implementation of above approach

C++

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// C++ Program to find MDAS_Factorial
#include <bits/stdc++.h>
using namespace std;
  
// Program to find MDAS_factorial
int MDAS_Factorial(int N)
{
    if (N <= 2)
        return N;
  
    if (N <= 4)
        return (N + 3);
  
    if ((N - 4) % 4 == 0)
        return (N + 1);
  
    else if ((N - 4) % 4 <= 2)
        return (N + 2);
  
    else
        return (N - 1);
}
  
// Driver code
int main()
{
  
    int N = 4;
    cout << MDAS_Factorial(N) << endl;
    N = 10;
    cout << MDAS_Factorial(N) << endl;
  
    return 0;
}

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Java

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// Java program find MDAS_Factorial
import java.util.*;
  
class Count {
    public static int MDAS_Factorial(int N)
    {
        if (N <= 2)
            return N;
  
        if (N <= 4)
            return (N + 3);
  
        if ((N - 4) % 4 == 0)
            return (N + 1);
  
        else if ((N - 4) % 4 <= 2)
            return (N + 2);
  
        else
            return (N - 1);
    }
  
    public static void main(String[] args)
    {
        int N = 4;
        System.out.println(MDAS_Factorial(N));
  
        N = 10;
        System.out.println(MDAS_Factorial(N));
    }
}

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Python3

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# Python3 code find MDAS_Factorial
def MDAS_Factorial( N ):
      
    if N <= 2:
        return N
  
    if N <= 4:
        return N + 3
          
    if (N - 4) % 4 == 0:
        return N + 1
  
    elif (N - 4) % 4 <= 2:
         return N + 2
  
    else:
         return N - 1
  
# Driver code
N = 4  
print(MDAS_Factorial( N ) )
  
N = 10 
print(MDAS_Factorial( N ) )

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C#

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// C# program to find MDAS_Factorial
using System;
  
class Count {
    public static int MDAS_Factorial(int N)
    {
        if (N <= 2)
            return N;
  
        if (N <= 4)
            return (N + 3);
  
        if ((N - 4) % 4 == 0)
            return (N + 1);
  
        else if ((N - 4) % 4 <= 2)
            return (N + 2);
  
        else
            return (N - 1);
    }
  
    // Driver code
    public static void Main()
    {
        int N = 4;
        Console.WriteLine(MDAS_Factorial(N));
  
        N = 10;
        Console.WriteLine(MDAS_Factorial(N));
    }
}

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PHP

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<?php
// PHP Program 
// Program to find MDAS_factorial
function MDAS_Factorial($N)
{
    if ($N <= 2)
          return N;
  
    if ($N <= 4)
          return ($N + 3);
  
    if (($N - 4) % 4 == 0)
          return ($N + 1);
  
    else if (($N - 4) % 4 <= 2)
          return ($N + 2);
  
    else
          return ($N - 1);
}
  
// Driver code
$N  = 4;
echo MDAS_Factorial($N);
echo("\n");
$N  = 10;
echo MDAS_Factorial($N);
?>

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Output :

7
12

Time complexity: O(1)
Auxiliary space: O(1)



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