Interview Question

Software Engineer Interview(Student Candidate)

How would you multiply two strings: "123 * "45", without

  any casting.
Answer

Interview Answer

9 Answers

0

I tried many methods but was not successful or/and efficient

Interview Candidate on Feb 28, 2014
0

How about getting the ascii code for each character, substracting 48, then you would have each digit. Then you can multiply them?

Alejandro on Mar 6, 2014
0

1. Add using a loop.

Or

2. Bitwise operation.

Anonymous on Mar 16, 2014
1

// C#

    static int StrToInt( string s )
    {
        if( s == null ) return int.MinValue;

        int n = 0;
        for( int i = s.Length - 1; i >= 0; --i )
            n = (n * 10) + (s[i] - '0');
        return n;
    }

    static int MultiplyStrings( string s1, string s2 )
    {
        int n1 = StrToInt( s1 );
        int n2 = StrToInt( s2 );
        return n1 * n2 ;
    }

Anonymous on Apr 2, 2014
1

public String multiply(String one, String two)
  {

      char[] one_char;
      char[] two_char;
      int carry = 0;

      if(one.length() > two.length()){

          one_char = one.toCharArray();
          two_char = two.toCharArray();
      }
      else{

          one_char = two.toCharArray();
          two_char = one.toCharArray();

      }

      int[] result = new int[one_char.length + two_char.length];

      for(int index = two_char.length - 1; index >= 0 ;index--){

          for(int index2 = one_char.length - 1; index2 >= 0 ;index2--){

              int value = ((one_char[index2] - '0') * (two_char[index] - '0')) + result[(index2 + index) + 1];
              result[(index2 + index) + 1] = value % 10;
              carry = value / 10;
              result[(index2 + index)] += carry;

          }
      }

     return Arrays.toString(result);

    }

Roshan on May 30, 2014
0

public static void main(String[] args) {

        String s1 = "4567";
        String s2 = "1234";

        int res = convertString2Int(s1) * convertString2Int(s2);

        System.out.println(res);

    }

    public static int convertString2Int(String s){

        char []chs = s.toCharArray();

        int n1 = 0;

        for(int i=0; i < chs.length ; i++){
            n1 = n1 + (chs[i]-'0') * getPowerOf(10, chs.length-i);
        }
        return n1;
    }

    public static int getPowerOf(int num, int times){

        int val = 1;
        for(int i =0; i<times-1; i++){
            val = val * num;
        }
        return val;
    }

jags on Aug 4, 2014
6

God damn these commentors are idiots..it clearly says that you cannot convert to an int and you paste pieces of memorized code where you are converting to int..smh

Jackasses on Aug 5, 2014
0

Subtracting char '0' is also what I first thought of but, If the restriction is not using any conversion to integers then one possible way would be to code up multiplication and addition tables (e.g., a dictionary of dictionaries would allow fast indexing into the result) and then using the algorithm from elementary school (or, if you feel really fancy, do something like Karatsuba multiplication http://en.wikipedia.org/wiki/Karatsuba_algorithm)

Eusebio on Oct 3, 2014
0

//Assumption +ve interger ... no sigh, no .,no float resut is within range of int
int multiplicationWithoutCast(string str1,string str2)
{
    int finalresult=0;

    if(str2.length() == 1)
    {
        int result=0;
        int intResult =0;
        int carry = 0;

        int j = 1;
        // used unsigned int i so infinite loop bcz never goes below zero
        for(int i=str1.length()-1; i>=0; i--,j=j*10) // u did j+10 first :/
        {
            intResult = (str1[i] - '0' ) * (str2[0] -'0') + carry;
            //cout << intResult << " ";
            carry = intResult/10;
            intResult = (intResult%10) * j;
            result = intResult + result;

            //cout << carry << " " << intResult << " " << result << endl;
        }

        if(carry)
            result = result + carry * j;

        return result;
    }
    else
    {
        for(int i=str2.length()-1,j=1;i>=0;i--,j=j*10)
        {
            finalresult = finalresult + multiplicationWithoutCast(str1,str2.substr(i,1)) * j;
        }

        return finalresult;
    }
}

Sandesh on Nov 18, 2014

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