Microsoft Interview Question for Software Engineers


Country: United States




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3
of 5 vote

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SOLUTION:

int smallestFactors(int n) {
    
if (n < 10) {
   
     return n;
    
}
    
List<Integer> list = new ArrayList<>();
   
     for (int i = 9; i > 1; i--) {
        
	while ((n % i) == 0) {
            
		n /= i;

          list.add(i);
 
       }

    }

    if (n > 10) {

        return 0;

    }
    
    int result = 0;
    
    for (int i : list) {

        result = result * 10 + i;
    
    }

    return result;

}

followup: what if we need to worry about the integer overflow?

- aonecoding June 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

I just tested this. This works perfectly. Other solutions are giving 94 and 98 as answers. This solution correctly gets you 49 and 89.

static int smallestFactors(int n)
        {
            if (n < 10)
            {
                return n;
            }

            List<int> list = new List<int>();

            for (int i = 9; i > 1; i--)
            {
                while ((n % i) == 0)
                {
                    n /= i;

                    list.Add(i);
                }
            }

            if (n > 10)
            {
                return 0;
            }

            int result = 0;
            list.Reverse();
            
            foreach (var i in list)
            {
                result = result * 10 + i;
            }

            return result;
        }

- Chander Dhall July 08, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

This is the correct solution (in C#)

static int smallestFactors(int n)
        {
            if (n < 10)
            {
                return n;
            }

            List<int> list = new List<int>();

            for (int i = 9; i > 1; i--)
            {
                while ((n % i) == 0)
                {
                    n /= i;

                    list.Add(i);
                }
            }

            if (n > 10)
            {
                return 0;
            }

            int result = 0;
            list.Reverse();
            
            foreach (var i in list)
            {
                result = result * 10 + i;
            }

            return result;

}

This is the correct solution in Python 3

def smallestFactors(n):
    list = []
    if n < 10:
        return n
    for i in range(9, 1, -1):
        while 0 == n % i:
            n = n / i
            list.append(i)
    if n > 10:
        return 0
    result = 0
    for i in reversed(list):
        result = result * 10 + i;
    return result

value = smallestFactors(36)

print("the answer is ", value);
value = smallestFactors(72)

print("the answer is ", value);

- me@chanderdhall.com July 08, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

unction smallest_factors(n) {
  let result = `no result found for ${n}`, 
      product;

  if( n < 10 ) return n;

  for(let i=+n+1; i<Number.MAX_SAFE_INTEGER; i++) {
    const sa = (''+i).split('');
    //console.log(sa);
    product = sa.reduce((acc,cur) => {
      //console.log(`entered reduce: acc=${acc}, cur=${cur}`);
      acc *= +cur;
      //console.log(`leaving reduce: acc=${acc}, cur=${cur}`);
      return acc;
    }, 1);

    //console.log(`product=${product}`);
    if( n == product ) {
      result = i;
      break;
    }
  }
  return result;
}

- dr. hfuhruhurr June 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

unction smallest_factors(n) {
  let result = `no result found for ${n}`, 
      product;

  if( n < 10 ) return n;

  for(let i=+n+1; i<Number.MAX_SAFE_INTEGER; i++) {
    const sa = (''+i).split('');
    product = sa.reduce((acc,cur) => {
      acc *= +cur;
      return acc;
    }, 1);

    if( n == product ) {
      result = i;
      break;
    }
  }
  return result;
}

- dr. hfuhruhurr June 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for n = 36, shouldn't be m = 49 instead of 66 ?

- manjeet.rulhania June 06, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.*;
import java.lang.*;
import java.io.*;
class test
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
                for(int i=n+1;i<2*n;i++){
                      int result=1;
                      int t=i;
                     while(t>0){
                        result=result*(t%10);
                        t=t/10;
                             }
                       if(result==n){
                        System.out.println(i);
                        break;}


                }
            
	}
}

if numbers are prime it will not print any thing.

- Anonymous June 07, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

for n = 36, m = 66 is wrong. m = 49 is smaller than 66.

- Anonymous June 07, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def get_smallest_int_made_of_factors(num):
  factors = []
  for digit in xrange(9, 1, -1):
    while num % digit == 0:
      factors.append(digit)
      num /= digit
  return int("".join([str(digit) for digit in reversed(factors)])) if num == 1 else None

- pasio June 10, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Decompose n to its prime factors. Sort the factors.

- Anonymous June 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

What should be answer for n=26

- Suresh June 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

what should be answer for n=26

- Suresh June 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 6 vote

Divide the number recursively with the largest digit possible (9,8,...,2). Save the digit at each step.

int reverse(int& value)
{
    int reverseNum = 0;
    while(value)
    {
        int digit = value % 10;
        reverseNum = reverseNum*10 + digit;
        value = value/10;
    }
    return reverseNum;
}

int main(int argc, const char * argv[]) {
    int number=27, answer = 0;
    int i=9;
    while(i > 1)
    {
        while(number%i == 0)
        {
            answer = answer*10 + i;
            number = number/i;
        }
        i--;
    }
    
    if(number > 1)
        cout<<"Not feasible";
    else
        cout<<reverse(answer); //reverse the digits
    return 0;
}

- Brandy June 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

static int smallestFactors(int n)
        {
            if (n < 10)
            {
                return n;
            }
            
            int result = 0, mul = 1;
            for (int i = 9; i > 1; i--)
            {
                while ((n % i) == 0)
                {
                    n /= i;

                    result = result == 0 ? i : i * mul + result;
                    mul = mul * 10;
                }
            }

            if (n > 10)
            {
                return 0;
            }            

            return result;
        }

- Jahnavi September 06, 2018 | Flag Reply


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