Implement a base 3 adder which takes two strings as input and returns a string

Question

donutello · Accepted Answer

base3DigitAdd(char lhs, char rhs,  char &extra, char &carry,char &sum)
{
  int s = (lhs - '0') + (rhs - '0') + (extra - '0');
  carry = (s/3) + '0';
  sum = (s%3) + '0';
}

base3Add(const string &lhs, const string &rhs, string &result)
{
  string::const_iterator il = lhs.end();
  string::const_iterator ir = rhs.end();
  result.clear();
  
  char carry = '0';
  char digit = '0';

  while((il != lhs.begin()) || (ir != rhs.begin())
  {
    char l = (il != lhs.begin) ? (*(il - 1)) : '0';
    char r = (ir != rhs.begin) ? (*(ir - 1)) : '0';

    base3DigitAdd(l, r, carry, carry, digit);
    result.push_front(digit);
   
    if (il != lhs.begin())
    {
      --il;
    }
    if (ir != rhs.begin())
    {
      --ir;
    }
  }

  if (carry != '0')
  {
    result.push_front(carry);
  }
}

Asa · Answer

string base3Add(const string & in1, const string & in2) {
	int i1 = atoi(in1.c_str());
	int i2 = atoi(in2.c_str());
	int carry = 0;
	char buff[2];
	string result;

	while(i1 != 0 || i2 != 0 || carry != 0) {
		int i1p = i1 % 10;
		int i2p = i2 % 10;
		int total = i1p + i2p + carry;
		itoa(total % 3, buff, 10);
		result.push_back(buff[0]);
		carry = total / 3;
		i1 /= 10;
		i2 /= 10;
	}

	reverse(result.begin(), result.end());

	return result;
}

Steve M · Answer

// sum : null terminated pre-allocated large enough string
char *addBase3(const char *a, const char *b, char *sum)
{
   char oldsum[20] = "";  // stacking characters would be much better than this
   const int alen = strlen(a);
   const int blen = strlen(b);

   int retainer = 0;
   int apos, aval;
   int bpos, bval;
   int partsum;
   int pos;

   for(pos = 1; pos = 3)
      {
         retainer = 1;
         partsum = 3 - partsum;
      }
      else
      {
         retainer = 0;
      }

      strcpy(oldsum, sum);
      sprintf(sum, "%d%s", partsum, oldsum);
   }

   if (retainer)
   {
      strcpy(oldsum, sum);
      sprintf(sum, "%d%s", retainer, oldsum);
   }

   return sum;
}

Steve M · Answer

if allowed, converting base 3 to base 2, then adding and converting back is better

Steve M · Answer

// sum : null terminated pre-allocated large enough string
char *addBase3(const char *a, const char *b, char *sum)
{
   char oldsum[20] = "";  // stacking characters would be much better than this
   const int alen = strlen(a);
   const int blen = strlen(b);

   int retainer = 0;
   int apos, aval;
   int bpos, bval;
   int partsum;
   int pos;

   for(pos = 1; pos = 3)
      {
         retainer = 1;
         partsum = 3 - partsum;
      }
      else
      {
         retainer = 0;
      }

      strcpy(oldsum, sum);
      sprintf(sum, "%d%s", partsum, oldsum);
   }

   if (retainer)
   {
      strcpy(oldsum, sum);
      sprintf(sum, "%d%s", retainer, oldsum);
   }

   return sum;
}

Andi Mullaraj · Answer

public class Test1 {

	public static void main(String[] args) {
		System.out.println(new Test1().add("21", "22")); //prints 120
	}

	int toInt(String str) throws NumberFormatException {
		int result = 0;
		for (int index = 0; index  2) {
				throw new NumberFormatException();
			}
			result = result * 3 + i;
		}

		return result;
	}

	String toStr(int i) throws NumberFormatException {
		StringBuffer sb = new StringBuffer();
		while (i > 0) {
			sb.insert(0, i % 3);
			i /= 3;
		}
		return sb.length() == 0 ? "0" : sb.toString();
	}

	String add(String str1, String str2) throws NumberFormatException {
		return toStr(toInt(str1) + toInt(str2));
	}
}

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