HackerRank Extra Long Factorials Problem Solution

In this post, We will solve HackerRank Extra Long Factorials Problem Solution.

The factorial of the integer n, written n!, is defined as:
n! = n x (n-1) x (n-2) x x3x2 x 1
Calculate and print the factorial of a given integer.
For example, if n = 30, we calculate 30 x 29 x 28 x … x 2 x 1 and get 265252859812191058636308480000000.
Function Description
Complete the extraLongFactorials function in the editor below. It should print the result and return.
extraLongFactorials has the following parameter(s):

n: an integer

Note: Factorials of n > 20 can’t be stored even in a 64-bit long long variable. Big integers must be used for such calculations. Languages like Java, Python, Ruby etc. can handle big integers, but we need to write additional code in C/C++ to handle huge values. We recommend solving this challenge using BigIntegers.

Input Format
Input consists of a single integer n
Constraints
1 ≤ n ≤ 100
Output Format
Print the factorial of n.
Sample Input
25
Sample Output
15511210043330985984000000
Explanation
25! 25 x 24 x 23 x x 3 x 2 x 1

HackerRank Extra Long Factorials Problem Solution

Extra Long Factorials C Solution

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

#define MaxC    200

char O[MaxC];
char S[MaxC];
char T[MaxC];

void Asd(int d1, int d2, int c1, char *o, int *c2) {
    int Sum;
    
    Sum = d1 + d2 + c1;
    
    *o = (Sum % 10) + 0x30;
    *c2 = (Sum / 10);
}

void Msd(int d1, int d2, int c1, char *o, int *c2) {
    int Sum;
    
    Sum = (d1 * d2) + c1;
    
    *o = (Sum % 10) + 0x30;
    *c2 = (Sum / 10);
}

void DoFactorial(int i) {
    int o, c;
    
    if (i == 100) {
        o = 2;
        while (o < MaxC-2) {
            O[o-2] = O[o];
            o++;
        }
        O[MaxC-3] = '0';
        O[MaxC-2] = '0';
    }
    else {     
        o = MaxC-2;
        c = 0;
        while (O[o] != ' ') {
            Msd(O[o]-0x30, i%10, c, S+o, &c);
            o--;
        }
    
        if (c > 0) S[o] = c + 0x30;
    
        if (i>9) {
            o = MaxC-2;
            c = 0;
            while(O[o] != ' ') {
                Msd(O[o]-0x30, i/10, c, T+o-1, &c);
                o--;
            }
    
            if (c > 0) T[o-1] = c + 0x30;
        }
    
        o = MaxC-2;
        c = 0;
        while ((S[o] != ' ') || (T[o] != ' ')) {
            if (S[o] == ' ')
                Asd(0, T[o]-0x30, c, O+o, &c);
            else
                if (T[o] == ' ')
                    Asd(S[o]-0x30, 0, c, O+o, &c);
                else
                    Asd(S[o]-0x30, T[o]-0x30, c, O+o, &c);
            o--;
        }
    
        if (c > 0) O[o] = c + 0x30;
    
    }
}

int main() {
    int n, i;
    
    for (i=0; i<MaxC-2; i++) {
        O[i] = ' ';
        S[i] = ' ';
        T[i] = ' ';
    }
    O[MaxC-2] = '1';
    S[MaxC-2] = '0';
    T[MaxC-2] = '0';
    O[MaxC-1] = 0;
    S[MaxC-1] = 0;
    T[MaxC-1] = 0;
    
    scanf("%d", &n);
    
    for (i=1; i<=n; i++) {
        DoFactorial(i);
    }
    
    i=0;
    while (O[i] == ' ') {
        i++;
    }    
    printf("%s", O+i);
    
 #if 0  
    i=0;
    while (S[i] == ' ') {
        i++;
    } 
    printf("\n%s", S+i);
    
    i=0;
    while (T[i] == ' ') {
        i++;
    }
    printf("\n%s", T+i); 
#endif
    
    return 0;
}

Extra Long Factorials C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    int number[300] = {};
    int n = 0;
    cin >> n;
    number[0] = 1;
    int len = 1;
    for (int i = 2; i <= n; ++i) {
        for (int pos = 0, r = 0; pos < len || r != 0; ++pos) {
            if (pos < len)
                r += number[pos] * i;
            else
                ++len;
            number[pos] = r % 10;
            r /= 10;
        }
    }
    for (int i = len - 1; i >= 0; --i)
        cout << number[i];
    return 0;
}

Extra Long Factorials C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
using System.Text;
using System.Linq;
class Solution {
    
      private static string Add(string first, string second)
        {
            var split = 0;
            if (first.Length < second.Length)
            {
                var s = new StringBuilder();
                s.Append('0', second.Length - first.Length);
                s.Append(first);
                first = s.ToString();
            }
            if (second.Length < first.Length)
            {
                var s = new StringBuilder();
                s.Append('0', first.Length - second.Length);
                s.Append(second);
                second = s.ToString();
            }
            var ret = new StringBuilder();

            for (int i = first.Length - 1; i >= 0; i--)
            {
                var rr = int.Parse(first[i].ToString()) + int.Parse(second[i].ToString()) + split;

                split = int.Parse(rr.ToString("00")[0].ToString());
                ret.Append(rr.ToString("00")[1]);
            }

            ret.Append(split);

            return new string(ret.ToString().ToCharArray().Reverse().ToArray());
        }

        private static string mul(string first, string second)
        {
            var ret = new string('0', first.Length);

            for (int i = second.Length - 1; i >= 0; i--)
            {
                var b = new StringBuilder();
                b.Append(mul(first, int.Parse(second[i].ToString())));
                b.Append('0', second.Length - 1 - i);

                ret = Add(ret, b.ToString());
            }
            var r = new StringBuilder();


            return ret.TrimStart('0');

        }

        private static string mul(string first, int second)
        {

            var split = 0;

            var ret = new StringBuilder();

            for (int i = first.Length-1; i >= 0; i--)
            {
                var rr = int.Parse(first[i].ToString()) * second+split;

                split = int.Parse(rr.ToString("00")[0].ToString());
                ret.Append(rr.ToString("00")[1]);
            }

            ret.Append(split);

            return new string(ret.ToString().ToCharArray().Reverse().ToArray());
        }    
    static void Main(String[] args) {
             var inp = int.Parse(Console.ReadLine());

            string ret = inp.ToString();
            for (int i = inp - 1; i > 0; i--)
            {
                ret = mul(ret, i.ToString());
                //Console.WriteLine(ret);
                
            }

            Console.WriteLine(ret);
        
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
    }
}

Extra Long Factorials Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
import java.math.BigInteger;

class Result {

    /*
     * Complete the 'extraLongFactorials' function below.
     *
     * The function accepts INTEGER n as parameter.
     */

    public static void extraLongFactorials(int n) {
        BigInteger sum=new BigInteger("1");
    for(int i=1;i<n+1;i++){
        sum=sum.multiply(BigInteger.valueOf(i));
    }
    System.out.println(sum);

    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        Result.extraLongFactorials(n);

        bufferedReader.close();
    }
}

Extra Long Factorials JavaScript Solution

function processData(input) {
    //Enter your code here
    input
    var res = 1
    for (var i = 1; i <= input; i++) {
        res = mult(res.toString(), i.toString())
    }
    process.stdout.write(res)
    function mult(num1,num2){
        var a1 = num1.split("").reverse();
        var a2 = num2.split("").reverse();
        var aResult = new Array;

        for ( iterNum1 = 0; iterNum1 < a1.length; iterNum1++ ) {
            for ( iterNum2 = 0; iterNum2 < a2.length; iterNum2++ ) {
                idxIter = iterNum1 + iterNum2;	// Get the current array position.
                aResult[idxIter] = a1[iterNum1] * a2[iterNum2] + ( idxIter >= aResult.length ? 0 : aResult[idxIter] );

                if ( aResult[idxIter] > 9 ) {	// Carrying
                    aResult[idxIter + 1] = Math.floor( aResult[idxIter] / 10 ) + ( idxIter + 1 >= aResult.length ? 0 : aResult[idxIter + 1] );
                    aResult[idxIter] -= Math.floor( aResult[idxIter] / 10 ) * 10;
                }
            }
        }
        return aResult.reverse().join("");
    }

} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Extra Long Factorials Python Solution

import math

case = int(input())

print(math.factorial(case))