CareerCup

  • Questions
  • Forum
  • Resume Tips
  • RSS
  • Sign In

Microsoft Interview Question for Software Engineer in Tests


More Questions from This Interview



Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is my solution, although I don't like it very much...

int digit(char* str, const char num, const char ten, const char five, const char one)  {  
    switch (num) {
    case 0:
        return 0;
        break;
    case 1:
        *str = one;
        return 1;
        break;
    case 2:
        *(str++) = one;
        *(str++) = one;
        return 2;
        break;
    ...
    ...
    case 6:
        *(str++) = five;
        *(str++) = one;
        return 2;
    ...   
    ...
    case 9:  
        *(str++) = one;
        *(str++) = ten;
        return 2;
    }
}

int convert(int number, char *buf) {
    if (!buf) return -1;

    if (number > 3999) return 0;  // don't have enough knowledge

    int the_d = number / 1000;
    buf += digit(buf, the_d, '?', '?', 'M');

    the_d = (number / 100) % 10;
    buf += digit(buf, the_d, 'M', 'D', 'C');

    the_d = (number / 10) % 10;
    buf += digit(buf, the_d, 'C', 'L', 'X');

    the_d = number % 10;
    buf += digit(buf, the_d, 'X', 'V', 'I');
  
     *buf = '\0';
    return 1;
}

- September 25, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

OK. One mistake. The signature of the first function should be:
int digit(char* str, int num, const char ten, const char five, const char one)

- Jason September 25, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

explain the question in detail please?

- kiran September 26, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

real roman to numeral is different that asked. is it purposfully or its mistake in question?

http://en.wikipedia.org/wiki/Roman_numbers

- raj September 26, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

can i represent 3999 = iMMMM?

how come 3999 = MMMCMXCIX?

Comment please?

- raj September 26, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 0 votes

The expression is based on digit.
3000=MMM
900=CM
90=XC
9=IX

I guess you cannot combine digits. So you cannot use iMMMM

- lifengg September 29, 2008 | Flag
Comment hidden because of low score. Click to expand.
0
of 2 vote

#include <stdio.h>
#include <iostream>

const char *Number_AsRomanString( int iNumber )
{
struct RomanDigit_t
  {
char *m_psString;
int m_iValue;
  };

static const RomanDigit_t RomanDigits[]=
  {
    {"M",  1000},
    {"CM",  900},
    {"D",   500},
    {"CD",  400},
    {"C",   100},
    {"XC",   90},
    {"L",    50},
    {"XL",   40},
    {"X",    10},
    {"IX",    9},
    {"V",     5},
    {"IV",    4},
    {"I",     1},
  };

// Strictly speaking, Roman digits can't display something
// such as 4999 without using overlaid bars and so forth,
// but for now this is a quick-and-dirty piece of code that'll
// just keep using M's...
//
static char sRomanString[20];
sRomanString[0] = '\0';

for (int i=0; iNumber && i<sizeof(RomanDigits)/
                           sizeof(RomanDigits[0]); i++)
  {
while ( RomanDigits[i].m_iValue <= iNumber )
    {
strcat( sRomanString, RomanDigits[i].m_psString );
iNumber -= RomanDigits[i].m_iValue;
    }
  }

return sRomanString;
}

int main(){
	int num = 16;
	const char *str = Number_AsRomanString(num);
	printf("\nThe number is %s",str);
}

- Laks September 26, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

can u explain the above code..specially the for loop..

- pm October 28, 2008 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

int findLen(int n)
{
int count = 0;
while(n)
{
switch(n%10)
{
case 1:
case 5:
count++; break; //I,V

case 2:
case 4:
case 6:
case 9;
count+=2; break; //II, IV, VI, IX

case 3:
count+=3; break; //III, VII

case 4:
count+=4; break; //VIII
}
n/=10;
}
return count;
}

void create(int num, int str)
{
char roman[4][] = {{'I', 'V'}, {'X', 'L'}, {'C', 'D'}, {'M'}};
char *ptr = str;
int rem;
int i = 0;
while(num)
{
switch(num%10)
{
case 3:
*ptr++ = roman[i][0];
case 2:
*ptr++ = roman[i][0];
case 1:
*ptr++ = roman[i][0]; break;
case 4:
*ptr++ = roman[i][0];
case 5:
*ptr++ = roman[i][1]; break;
case 6:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; break;
case 7:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0];
break;
case 8:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0]; break;
break;
case 9:
*ptr++ = roman[i][0]; *ptr++ = roman[i+1][0]; break;
}
n/=10;
}
strrev(str);
}

int main()
{
char *str = NULL;
len = findLen(num) + 1;
str = (char*)malloc(sizeof(char) * len);
create(num, str);
return 0;
}

- mail2vcp@gmail.com September 30, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int findLen(int n)
{
int count = 0;
while(n)
{
switch(n%10)
{
case 1:
case 5:
count++; break; //I,V

case 2:
case 4:
case 6:
case 9;
count+=2; break; //II, IV, VI, IX

case 3:
count+=3; break; //III, VII

case 8:////////////////////////////////////mistake corrected
count+=4; break; //VIII
}
n/=10;
}
return count;
}

void create(int num, int str)
{
char roman[4][] = {{'I', 'V'}, {'X', 'L'}, {'C', 'D'}, {'M'}};
char *ptr = str;
int rem;
int i = 0;
while(num)
{
switch(num%10)
{
case 3:
*ptr++ = roman[i][0];
case 2:
*ptr++ = roman[i][0];
case 1:
*ptr++ = roman[i][0]; break;
case 4:
*ptr++ = roman[i][0];
case 5:
*ptr++ = roman[i][1]; break;
case 6:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; break;
case 7:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0]; 
break;
case 8:
*ptr++ = roman[i][1]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0]; *ptr++ = roman[i][0]; break;
break;
case 9:
*ptr++ = roman[i][0]; *ptr++ = roman[i+1][0]; break;
}
n/=10;
}
strrev(str);
}

int main()
{
char *str = NULL;
len = findLen(num) + 1;
str = (char*)malloc(sizeof(char) * len);
create(num, str);
return 0;
}

- mail2vcp@gmail.com; small mistake September 30, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

can you explain the above code..could u explain how will it work for 45..in the above code where is i incremented? n wats the role of n?

- ad October 04, 2008 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

string convert(int number){
	string* numerals = new string[7];
	numerals[0] = "I";
	numerals[1] = "V";
	numerals[2] = "X";
	numerals[3] = "L";
	numerals[4] = "C";
	numerals[5] = "D";
	numerals[6] = "M";
	
	int order = 0;
	int current = number;
	string roman = ""; 
	
	while( number != 0 ){
		current = number % 10;
		number = number / 10;
		
		if(current <= 3){
			for(int r=0; r<current; r++)
				roman = numerals[2*order] + roman;
		}
		else if(current == 4)
			roman = numerals[2*order]+numerals[2*order+1] + roman;
		else if(current == 5)
			roman = numerals[2*order+1] + roman;
		else if(current <= 8){
			for(int r=0; r<(current-5); r++)
				roman = numerals[2*order] + roman;
			roman = numerals[2*order+1] + roman;
		}
		else if(current == 9)
			roman = numerals[2*order]+numerals[2*order+2] + roman;
		
		order++;
		if(order > 6)
			throw(-1);
	}
	return roman;
}

- January 20, 2009 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

//Use Recursion to solve it

char roman[] = "IVXLCDM";

void Int_To_Roman(int number,char *str)
{
	int digit = number % 10;
	int mod = digit%5;
	if(mod<=3)
	{
		for(int i=0;i<mod;i++)
		{
			*str++ = roman[index];
		}
		if(digit/5>=1)
		{
			*str++ = roman[index+1];
		}
	}
	else
	{
		if(digit/5>=1)
		{
		*str++ = roman[index+2];
		}
		else
		{
		*str++ = roman[index+1];
		}
		*str++ = roman[index];
	}
	*str = '\0';
	index += 2;

	if(number/10>0)
	{
		Int_To_Roman(number/10,str);
	}
}

- Arun June 15, 2009 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

string to_roman(int n){
	if(n>=4000)
		return "TOO LARGE INPUT";
	map<int,char> roman_numerals;
	roman_numerals[1000]='M';roman_numerals[500]='D';roman_numerals[100]='C';roman_numerals[50]='L';roman_numerals[10]='X';roman_numerals[5]='V';roman_numerals[1]='I';	
	string result="";
	for(int place=1000;place>0;place/=10){
		int face=n/place;
		if (face<=3)
			for(int i=0;i<face;i++)
				result.push_back(roman_numerals[place]);
		else if(face==4){
			result.push_back(roman_numerals[place]);
			result.push_back(roman_numerals[place*5]);
		}
		else if (face==5)
			result.push_back(roman_numerals[place*5]);
		else if (face<=8){
			result.push_back(roman_numerals[place*5]);
			for(int i=6;i<=face;i++)
				result.push_back(roman_numerals[place]);
		}
		else if(face==9){
			result.push_back(roman_numerals[place]);
			result.push_back(roman_numerals[place*10]);
		}
		n%=place;
	}
	return result;
}

- Anonymous August 01, 2009 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

.

- Anonymous March 21, 2010 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int map(char inp)
{
switch (inp)
{
case 'i':
case 'I':
return 1;
case 'v':
case 'V':
return 5;
case 'x':
case 'X':
return 10;
case 'l':
case 'L':
return 50;
case 'c':
case 'C':
return 100;
case 'd':
case 'D':
return 500;
default:
return 0;
}
}


int main()
{
char str[16];
int res = 0,prev = 501,curr = 0;
cout <<"Enter the input ROMAN string = ";
cin >> str;
int len = strlen(str);
for(int i=0;i<len;i++)
{
curr = map(str[i]);
if(curr<=prev)
res += curr;
else
{
res += curr;
res -= 2*prev;
}
prev = curr; //Re-initialize the previous
}

cout <<"The decimal equivalent of input Roman "<<str<<" = "<<res;
return 0;
}

- Anonymous March 21, 2010 | Flag Reply


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More