Goldman Sachs Interview Question for Applications Developers


Country: India
Interview Type: Written Test




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3
of 3 vote

public class LargeMultiplication
{
public static void main( String[] args )
{
int[] Num1 = { 5,3,6,2,8,2,0,2,8 };
int[] Num2 = { 3,5,2,3,2,1,3,4 };
int sum = 0, carry;
int k = Num1.length + Num2.length;
int[] Num = new int[k];
for( int i = Num1.length - 1; i >= 0; i-- )
{
carry = 0;
for( int j = Num2.length - 1; j >= 0; j-- )
{
sum = Num2[j] * Num1[i] + carry + Num[k - 1];
carry = sum / 10;
Num[--k] = sum % 10;

}
Num[--k] = carry;
k += Num2.length;
}

for( int j = 0; j < Num.length; j++ )
{
System.out.println( j + "= " + Num[j] );
}
}
}

- Pradeep kumar R S January 15, 2015 | Flag Reply
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1
of 1 vote

Use arrays to store the numbers, then for multiplication you will need n no of arrays, where n is the no of digits of smaller number and one extra array to store carry ,
Then perform multiplication :)

- Rahul January 10, 2015 | Flag Reply
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0
of 0 votes

Carry can also be managed without any extra carry array.

- PKT January 11, 2015 | Flag
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0
of 0 votes

You also need to pad the array from the multiplication with second number onwards with zeros to get the correct sum result.

Ex:
123
46
-----
738
492"0"
--------
5658
--------

- Vib February 05, 2015 | Flag
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0
of 0 vote

BigInteger if you allowed to use built-in java classes

- Darkhan.Imangaliev January 11, 2015 | Flag Reply
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0
of 0 vote

use linked list

- DashDash January 11, 2015 | Flag Reply
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0
of 0 vote

C++ solution

Btw: The second example multiplication is wrong, it's 536282028 X 352321 (one less 8)

#include <algorithm>
#include <vector>
using namespace std;

vector<int> MultiplyVectors(const vector<int> &v1, const vector<int> &v2) {
  vector<vector<int>> sums;
  for (int i = v2.size()-1; i >= 0; --i) {
    sums.push_back(vector<int>(v2.size()-1-i, 0));
    int carry = 0;
    for (int j = v1.size()-1; j >= 0; --j) {
      int v = v1[j]*v2[i]+carry;
      sums.back().push_back(v%10);
      carry = v/10;
    }
    if (carry != 0)
      sums.back().push_back(carry);
  }

  size_t max_idx = 0;
  for (size_t i = 0; i < sums.size(); ++i)
    max_idx = max(max_idx, sums[i].size());

  vector<int> ret;
  int v = 0;
  for (size_t i = 0; i < max_idx; ++i) {
    for (size_t j = 0; j < sums.size(); ++j) {
      if (i < sums[j].size())
        v += sums[j][i];
    }
    ret.push_back(v%10);
    v /= 10;
  }
  if (v != 0)
    ret.push_back(v);

  reverse(ret.begin(), ret.end());
  return ret;
}

- Tobi January 14, 2015 | Flag Reply
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0
of 0 vote

#include<iostream>
using namespace std;

int main()
{
    int a[]={1,4,7,5,2,3};
    int n=sizeof(a)/sizeof(*a);

    int b[]={5,4,5,7,8,9,2};
    int m=sizeof(b)/sizeof(*b);

    int res[10000]={0};
    int carry=0;
    int h=0;
    for(int i=n-1; i>=0; i--)
    {
        int k=b[m-1]*a[i];
        k+=carry;
        res[h++]=k%10;
        carry=k/10;
    }

    while(carry)
    {
        res[h++]=carry%10;
        carry=carry/10;
    }


    int l=1;
    int k;
    for(int i=m-2; i>=0; i--)
    {
        int carry=0;
        k=l;
        int temp[1000]={0};

        for(int j=n-1; j>=0; j--)
        {
            int x=b[i]*a[j];
            x+=carry;
            carry=x/10;
            temp[k++]=x%10;
        }

        while(carry)
        {
            temp[k++]=carry%10;
            carry=carry/10;
        }

        for(int i=0; i<k; i++)
        {
            int sum=res[i]+temp[i];
            sum+=carry;
            carry=sum/10;
            res[i]=sum%10;
        }
        l++;
    }

    for(int i=k-1; i>=0; i--)
        cout<<res[i]<<" ";
}

- Saurabh October 01, 2015 | Flag Reply
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0
of 0 vote

#include<iostream>
using namespace std;

int main()
{
    int a[]={1,4,7,5,2,3};
    int n=sizeof(a)/sizeof(*a);

    int b[]={5,4,5,7,8,9,2};
    int m=sizeof(b)/sizeof(*b);

    int res[10000]={0};
    int carry=0;
    int h=0;
    for(int i=n-1; i>=0; i--)
    {
        int k=b[m-1]*a[i];
        k+=carry;
        res[h++]=k%10;
        carry=k/10;
    }

    while(carry)
    {
        res[h++]=carry%10;
        carry=carry/10;
    }


    int l=1;
    int k;
    for(int i=m-2; i>=0; i--)
    {
        int carry=0;
        k=l;
        int temp[1000]={0};

        for(int j=n-1; j>=0; j--)
        {
            int x=b[i]*a[j];
            x+=carry;
            carry=x/10;
            temp[k++]=x%10;
        }

        while(carry)
        {
            temp[k++]=carry%10;
            carry=carry/10;
        }

        for(int i=0; i<k; i++)
        {
            int sum=res[i]+temp[i];
            sum+=carry;
            carry=sum/10;
            res[i]=sum%10;
        }
        l++;
    }

    for(int i=k-1; i>=0; i--)
        cout<<res[i]<<" ";
}

- Saurabh October 01, 2015 | Flag Reply
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0
of 0 vote

public static void main(String[] args) {
int[] num1={1,2,3,4,5};
int[] num2={1,2,4,6};
System.out.println(multiply(breakNumbers(num1), breakNumbers(num2)));
int sum=0;
Map<Integer,Integer> multiplyMap = multiply(breakNumbers(num1), breakNumbers(num2));
for (Map.Entry<Integer,Integer> entry : multiplyMap.entrySet()) {
sum +=(entry.getKey()*(Math.pow(10,entry.getValue())));
}
System.out.println(sum);
}

static Map<Integer,Integer> multiply(Map<Integer,Integer> num1,Map<Integer,Integer> num2){
Map<Integer,Integer> multiplyMap=new LinkedHashMap<Integer,Integer>();
for (Map.Entry<Integer,Integer> entry1 : num1.entrySet()) {
for (Map.Entry<Integer,Integer> entry2 : num2.entrySet()) {
multiplyMap.put(entry1.getKey()*entry2.getKey(), entry1.getValue()+entry2.getValue());
}
}
return multiplyMap;
}

static Map<Integer,Integer> breakNumbers(int[] num1){
Map<Integer,Integer> numMap=new LinkedHashMap<Integer,Integer>();
String s="";
int count=0;
for(int i=0;i<num1.length;i++){
count++;
if(count==3 || i==num1.length-1){
s+=num1[i];
numMap.put(Integer.valueOf(s), num1.length-(i+1));
s="";
count=0;
}else{
s+=num1[i];
}

}
return numMap;
}

- prmd89 November 14, 2015 | Flag Reply
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0
of 0 vote

# Reverse an array to change endianness of numbers
def num_array_reverse(num_array):
        left = 0
        right = len(num_array) - 1;
        while (left < right):
                num_array[left], num_array[right] = num_array[right], num_array[left]
                left += 1
                right -= 1

# Multiply two numbers
# Input: Big-endian order
# Output: Big-endian order
def multiply(num_a_array, num_b_array):
        # Convert big-endian to little-endian
        num_array_reverse(num_a_array)
        num_array_reverse(num_b_array)

        # Initialize the result
        num_result_arr = []

        # Multiply
        for i in range(0, len(num_a_array)):
                carry = 0
                j = 0
                while (j < len(num_b_array)):
                        val = num_a_array[i] * num_b_array[j] + carry
                        if (len(num_result_arr) <= i + j):
                                num_result_arr.append(0)
                        val += num_result_arr[i + j]
                        carry = val / 10
                        val %= 10
                        num_result_arr[i + j] = val
                        j += 1
                if (carry):
                        if (len(num_result_arr) <= i + j):
                                num_result_arr.append(carry)
                        else:
                                num_result_arr[i + j] = carry

        #  Convert little-endian to big-endian
        num_array_reverse(num_a_array)
        num_array_reverse(num_b_array)
        num_array_reverse(num_result_arr)

        return num_result_arr

def multiply_and_print(num_a_array, num_b_array):
        num_result_array = multiply(num_a_array, num_b_array)
        print("{0} x {1} = {2}".format(num_a_array, num_b_array, num_result_array))

def main():
        multiply_and_print([1, 2], [1, 0])
        multiply_and_print([5, 3, 6, 2, 8, 0, 2, 8], [3, 5, 2, 3, 2, 1])

main()

- sw.engineers.com January 21, 2016 | Flag Reply


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