Microsoft Interview Question for SDETs


Team: Cloud
Country: United States




Comment hidden because of low score. Click to expand.
7
of 7 vote

int toInt(String str, int base){

		checkArgument(str != null, "'str' parameter is NULL");
		checkArgument(base > 1, "Incorrect 'base' passed");

		str = str.trim().toLowerCase();
		checkArgument(str.length() > 0, "Zero length/empty 'str' parameter passed");

		if( "-2147483648".equals(str) ){
			return Integer.MIN_VALUE;
		}

		int sign = 1;
		int index = 0;

		if( str.charAt(0) == '-' ){
			sign = -1;
			++index;
		}
		else if( str.charAt(0) == '+'){
			++index;
		}

		int res = 0;

		for( int i = index; i < str.length(); i++ ){
			char ch = str.charAt(i);

			int digit;

			if( Character.isDigit(ch) ){
				digit = ch - '0';
			}
			else {
				digit = 10 + (ch - 'a');
			}

			if( digit >= base ){
				throw new IllegalArgumentException("Incorrect str '" + str  +"', for base " + base);
			}

			res = res * base + digit;

			if( res < 0 ){
				throw new ArithmeticException("Overflow occurred for value '" + str + "'");
			}
		}

		return sign * res;
	}

- m@}{ January 24, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
8
of 8 votes

Coool m@}{ you provided the most complete solution !

- dumb fellow January 24, 2015 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

int ConvertStrToInt(string inputStr)
{
	int num = 0;
	char[] inputCharArray = inputStr.ToCharArray();
	foreach (char ch in inputCharArray)
	{
		num = num * 10 + (ch - '0');
	}
	return num;
}

- dumb fellow January 24, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

public static int ConvertToInteger(string number)
        {
            int sign = 1;
            int returnValue = 0;

            foreach (var digit in number)
            {
                if (digit == '+') continue;
                if (digit == '-')
                {
                    sign = -1;
                    continue;
                }

                if (digit < '0' || digit > '9')
                    throw new ArgumentException("String cannot have numbers!");

                returnValue = (returnValue * 10) + (digit - '0');
            }

            return returnValue * sign;
        }

- Asiful Haque Mahfuze February 10, 2015 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

private int parseNumber(String unparsedNumber) {
	final char[] unparsedNumberCharacters = unparsedNumber.toCharArray();

	int parsedNumber = 0;
	int currentPlace = 1;

	for (int i = (unparsedNumberCharacters.length - 1); i >= 0; i--) {
		parsedNumber += Character.getNumericValue(unparsedNumberCharacters[i]) * currentPlace;

		currentPlace *= 10;
	}

	return parsedNumber;
}

- Chris January 26, 2015 | Flag Reply
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0
of 0 vote

public class StringtoNumber {

	public static void main(String[] args){
		String num = "1234";
		int k =1;
		int sum =0;
		char ch[] = num.toCharArray();
		for(int i=ch.length-1;i>=0;i--){
			
			sum = sum + Character.getNumericValue(ch[i]) * k;
			k= k * 10;
			
		}
		System.out.println(sum);
	}

}

- umamahes January 26, 2015 | Flag Reply
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0
of 0 vote

int convert_to_num(char *arr)
{
   int num = 0;
   int len = 0;
   int i =1;
   int is_negative = 0;
   char *tmp = arr;

   if(*tmp)
   {

     while(*tmp++)  /* len of string*/
        len++;

     tmp = arr;
     if(*tmp == '-') {
         len = len -1;  /*  first char is a sign for -ve */
         is_negative =1;
         tmp = tmp + len;
     }
     else
     {
         tmp = tmp + (len -1);  /* To last char in the list */
     }

     while(len--) {
        if(('0' <= *tmp) && ('9' >= *tmp))
        {
            num = num + ((*tmp - '0')*i);
            tmp--;
            i=i*10;
        }
        else
        {
            return 0;
        }
     }
   }
   if(num && is_negative) {
        num = ~num+1;
   }
   return num;
}

- Anonymous January 27, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class stringToInteger {

	public static void main(String[] args) {
		int y=1;
		int number = 0;
		int start = 0;
		int sign =1;
		String str = "9000";
		
		if(str.charAt(0) == '+'){
			start++;
			sign = 1;
		}
		else if(str.charAt(0) == '-'){
			start++;
			sign = -1;
		}
		for(int i=str.length()-1; i>=start;i--){
			char c = str.charAt(i);
			number = number + ((c-48)*y);
			y*=10;
			
		}
		number *= sign;
		
		System.out.println(number);
	}
}

- Joey January 27, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
void convert(char* in){
int num=0,i=0;
while(in[i]!='\0'){
num=num*10+(in[i]-'0');
i++;
}
printf("converted to decimal %d",num);
}
void main(){
char str[10]="1234";
convert(str);
}

- Prashanth Artal (ASU) January 27, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Stringtointeger{
public static void main(String args[]){
String s="12323456";
System.out.println(Integer.parseInt(s));
}
}

- Anonymous February 18, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String args[]){


StringstoInt();

}

static void StringstoInt(){
String str = "123456";
int a;

int len = str.length();
//System.out.println(""+len);

for(int i=0;i<len;i++){
a = str.charAt(i) - 48;
//System.out.println(""+i);
System.out.print(""+a);

}

- Anonymous March 05, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String args[]){
		
		
		StringstoInt();
		
	}
	
	static void StringstoInt(){
		String str = "123456";
		int a;
		
		int len = str.length();
		//System.out.println(""+len);
		
		for(int i=0;i<len;i++){
		 a =  str.charAt(i) - 48;
			//System.out.println(""+i);
		System.out.print(""+a);

}

- sai March 05, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Assume if the input is not a numeric string then throw exception:

private int stringToInt(String s){
        if(isNumeric(s)) {
            long maxInt = Integer.MAX_VALUE;
            long minInt = Integer.MIN_VALUE;

            long stringValue = Long.parseLong(s);

            if (stringValue > maxInt || stringValue < minInt) {
                throw new ArithmeticException("String value : " + s + " is beyond integer range.");
            } else {
                return Integer.parseInt(s);
            }
        }
        else{
            throw new NumberFormatException("The String input is not valid.");
        }
    }

    private boolean isNumeric(String s){
        for(int i = 0; i<s.length(); ++i){
            try{
                Integer.parseInt(Character.toString(s.charAt(i)));
            }
            catch(NumberFormatException e){
                return false;
            }
        }
        return true;

}

- Anonymous March 11, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Well, since the question does not ask a particular way to solve just use int.TryParse

int ToInt(string input)
   int output;
    if( int.TryParse(input,out output))
      {
        return output;
      }
     throw new ArgumentException();
  }

- Daniel March 12, 2015 | Flag Reply


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