Facebook Interview Question for Data Engineers


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
4
of 4 vote

In Python 3

def string_multiplication(a, b):
    result = 0
    for i in range(len(a)):
        for j in range(len(b)):
            result += (ord(a[i])-ord('0'))*(10**(len(a)-i-1))*(ord(b[j])-ord('0'))*(10**(len(b)-j-1))
    return result

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234
print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Florent L July 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
2
of 2 vote

decimal multiplication on string does it, although converting the string to binary and doing a binary multiplication would obviously be more effective

in python 3

def string_multiplication(a, b):
    result = []
    result_start_idx = 0    
    for ai in map(lambda x: ord(x) - ord('0'), reversed(a)):
        overflow = 0
        result_idx = result_start_idx
        for b_idx in range(len(b) - 1, -1, -1):
            m = (ord(b[b_idx]) - ord('0')) * ai + overflow
            if result_idx < len(result): 
                m += result[result_idx]
                result[result_idx] = m % 10
            else :
                result.append(m % 10)
            overflow = m // 10
            result_idx += 1
        if overflow > 0:
            result.append(overflow)
        result_start_idx += 1
    return ''.join(map(lambda x: chr(ord('0') + x), reversed(result)))

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234
print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Chris July 01, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#the product of ith index of A and jth index of B would be going to i+j and i+j+1 th position
def string_multiplication(a, b):
def string_multiplication(a, b):
    result=[0]*(len(a)+len(b))
    for i in range(len(a)-1,-1,-1):
        for j in range(len(b)-1,-1,-1):
            mul=int(a[i])*int(b[j])
            pos1,pos2=i+j,i+j+1
            sum_all=mul+result[pos2]
            result[pos1]+=int(sum_all/10)
            result[pos2]=int(sum_all%10)
    return result
print(string_multiplication('8', '100'))

- Murali July 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):
    result=[0]*(len(a)+len(b))
    for i in range(len(a)-1,-1,-1):
        for j in range(len(b)-1,-1,-1):
            mul=int(a[i])*int(b[j])
            pos1,pos2=i+j,i+j+1
            sum_all=mul+result[pos2]
            result[pos1]+=int(sum_all/10)
            result[pos2]=int(sum_all%10)
    return result
print(string_multiplication('8', '100'))

- Anonymous July 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):
    result=[0]*(len(a)+len(b))
    for i in range(len(a)-1,-1,-1):
        for j in range(len(b)-1,-1,-1):
            mul=int(a[i])*int(b[j])
            pos1,pos2=i+j,i+j+1
            sum_all=mul+result[pos2]
            result[pos1]+=int(sum_all/10)
            result[pos2]=int(sum_all%10)
    return result
print(string_multiplication('8', '100'))

- Murali July 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def multiplystrings(s1,s2):
    totalsum,sum = 0,0
    for i in range(len(s1)):
        sum = 0
        for j in range(len(s2)):
            sum *= 10
            sum += int(s1[i]) * int(s2[j])
        totalsum *= 10
        totalsum += sum
    print(totalsum)

- Srujan July 09, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def multiplystrings(s1,s2):
totalsum,sum = 0,0
for i in range(len(s1)):
sum = 0
for j in range(len(s2)):
sum *= 10
sum += int(s1[i]) * int(s2[j])
totalsum *= 10
totalsum += sum
print(totalsum)

- Srujan July 09, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Same solution as given by Florent but in Java:

import java.util.*;
class Main {
    public static void main(String[] args) {
        String n1 = "3922445243";
        String n2 = "3248234";
       
        double result=0;
        for(int i=0;i<n1.length();i++){
           for(int j=0;j<n2.length();j++){
                double num1 = (n2.charAt(j)-'0')*(Math.pow(10,n2.length()-j-1));
                double num2 = (n1.charAt(i)-'0')*(Math.pow(10,n1.length()-i-1));
                result +=(num1*num2);
            }
        }
        System.out.println(result);
    }
}

- krupen.ghetiya July 27, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):

result = [0] * (len(a) + len(b))
for i in reversed(range(len(a))):
for j in reversed(range(len(b))):
result[i+j+1] += (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0'))
result[i+j] += result[i+j+1] / 10
result[i+j+1] %= 10

return ''.join( map(str, result[ next( idx for idx, digit in enumerate(result) if digit != 0) : ]) )

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234

print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Pradeep November 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):

    result = [0] * (len(a) + len(b))
    for i in reversed(range(len(a))):
        for j in reversed(range(len(b))):
            result[i+j+1] += (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0'))
            result[i+j]   += result[i+j+1] / 10
            result[i+j+1] %= 10

    return ''.join( map(str, result[ next( idx for idx, digit in enumerate(result) if digit != 0) : ]) )

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234

print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Pradeep November 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):

    result = [0] * (len(a) + len(b))
    for i in reversed(range(len(a))):
        for j in reversed(range(len(b))):
            result[i+j+1] += (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0'))
            result[i+j]   += result[i+j+1] / 10
            result[i+j+1] %= 10

    return ''.join( map(str, result[ next( idx for idx, digit in enumerate(result) if digit != 0) : ]) )

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234

print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Pradeep November 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def string_multiplication(a, b):
    result = [0] * (len(a) + len(b))
    for i in reversed(range(len(a))):
        for j in reversed(range(len(b))):
            result[i+j+1] += (ord(a[i]) - ord('0')) * (ord(b[j]) - ord('0'))
            result[i+j]   += result[i+j+1] / 10
            result[i+j+1] %= 10

    return ''.join( map(str, result[ next( idx for idx, digit in enumerate(result) if digit != 0) : ]) )

number_one = 193283492420348904832902348908239048823480823
number_two = 3248234890238902348823940990234
print(number_one*number_two)
print(string_multiplication(str(number_one), str(number_two)))

- Anonymous November 25, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def multiplication(s1, s2):
    product = []
    counter = 0
    for i in range(len(s2)-1,-1,-1):
        product.append(int(s2[i])*int(s1)*(10**counter))
        counter+=1
    return sum(product)

- Anonymous December 10, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

As an alternative solution, the code below uses recursion and dictionaries. It's more lines of code, but it also accounts for the negative edge case.

def multiplyStrings(str1, str2):
    
    if(str1[0]=="-"):
        initStr1 = str1[1:]
        multStr1 = -1
    else:
        initStr1 = str1[0:]
        multStr1 = 1
    
    if(str2[0]=="-"):
        initStr2 = str2[1:]
        multStr2 = -1
    else:
        initStr2 = str2[0:]
        multStr2 = 1
        
    return multStr1*multiplyString1(initStr1)*multStr2*multiplyString2(initStr2)
    

def multiplyString1(str1):
    
    if(len(str1)==1):
        return findNumber(str1)
    else:
        return findNumber(str1[0])*(10**(len(str1)-1))+multiplyString1(str1[1:])
    

def multiplyString2(str2):
    if(len(str2)==1):
        return findNumber(str2)
    else:
        return findNumber(str2[0])*(10**(len(str2)-1))+multiplyString2(str2[1:])


def findNumber(string):
    numbers = {
            "0" : 0,
            "1" : 1,
            "2" : 2,
            "3" : 3,
            "4" : 4,
            "5" : 5,
            "6" : 6,
            "7" : 7,
            "8" : 8,
            "9" : 9
            }
    
    return numbers[string]

num1 = 193283492420348904832902348908239048823480823
num2 = 3248234890238902348823940990234

print(num1*num2)

print(multiplyStrings(str(num1),str(num2)))

- Brent October 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def multiplyStrings(str1, str2):
    
    if(str1[0]=="-"):
        initStr1 = str1[1:]
        multStr1 = -1
    else:
        initStr1 = str1[0:]
        multStr1 = 1
    
    if(str2[0]=="-"):
        initStr2 = str2[1:]
        multStr2 = -1
    else:
        initStr2 = str2[0:]
        multStr2 = 1
        
    return multStr1*multiplyString1(initStr1)*multStr2*multiplyString2(initStr2)
    

def multiplyString1(str1):
    
    if(len(str1)==1):
        return findNumber(str1)
    else:
        return findNumber(str1[0])*(10**(len(str1)-1))+multiplyString1(str1[1:])
    

def multiplyString2(str2):
    if(len(str2)==1):
        return findNumber(str2)
    else:
        return findNumber(str2[0])*(10**(len(str2)-1))+multiplyString2(str2[1:])


def findNumber(string):
    numbers = {
            "0" : 0,
            "1" : 1,
            "2" : 2,
            "3" : 3,
            "4" : 4,
            "5" : 5,
            "6" : 6,
            "7" : 7,
            "8" : 8,
            "9" : 9
            }
    
    return numbers[string]

num1 = 193283492420348904832902348908239048823480823
num2 = 3248234890238902348823940990234

print(num1*num2)

print(multiplyStrings(str(num1),str(num2)))

- brent October 28, 2018 | 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