Adobe Interview Question for Member Technical Staffs


Country: India
Interview Type: In-Person




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2
of 2 vote

O(n)

1. Find the cube of each number starting from 1 to (n-3) and store in an array.
2. Search the combination of numbers which can sum to n ( I used binary search to do so)

// Use binary search to find the element.
int getIndex(int* data, int start, int end, int num)
{
    if ( start < end )
    {
        if ( data[(start + end)/2] == num )
        {
            return ((start + end)/2);
        }
        else if ( data[(start + end)/2] > num )
        {
            return getIndex(data,start,(start + end)/2, num);
        }
        else
        {
            return getIndex(data,(start + end)/2 + 1, end, num);
        }
    }
    else if ( start == end)
    {
        return data[start] == num ? start : -1;
    }
    return -1;
}


void printAllCombination(const int num)
{
    //std::map<int,int> cube;
    const int cubeArrsize = (num/3) + 1;
    int *cubeArr = new int[num + 1];

    //  Find the cube of each number and store in array.
    for ( int i = 1; i< cubeArrsize; i++)
    {
        cubeArr[i] = i*i*i;
    }

    // Do a binary search for the remaining number from the given position.
    for ( int i = 1; i< cubeArrsize; i++)
    {
        int j = getIndex(cubeArr, i+1, cubeArrsize,num - cubeArr[i]);
        if ( j > 0 )
        {
            std::cout<<"\n"<<i<<" , "<<j;
        }
    }
    delete[] cubeArr;
}

int main()
{
    printAllCombination(224);
    std::cout<<std::endl;
    getch();
}

- Ricky January 31, 2014 | Flag Reply
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0
of 0 votes

Why cubarrsize=size/3+1?

- aka January 31, 2014 | Flag
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0
of 0 votes

what about N = 0
a = -b
there are infinite pairs

- Anonymous February 01, 2014 | Flag
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0
of 0 votes

O(n) is funny.

- Anonymous February 14, 2014 | Flag
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0
of 0 votes

its O(nlogn) not O(n)

- desicoder May 31, 2014 | Flag
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0
of 0 vote

O(n) solution using a set

#include <iostream>
#include <set>
#include <cmath>

int main() {
	int N = 91;
	std::set<long> cubics_set;

	// Generating all cubic values between 1 and N
	for(int i = 1; i < N; i++) {
		cubics_set.insert(std::pow(i, 3));
	}

	for(auto it = cubics_set.begin(); it != cubics_set.end(); ++it) {
		if(*it > N)
			break;

		long required = N - *it;

		if(cubics_set.find(required) != cubics_set.end())
			std::cout << "a:" << *it << " "
				  << "b:" << required << std::endl;

	}

	return 0;
}

- Felipe Cerqueira January 30, 2014 | Flag Reply
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0
of 0 votes

am i loosing it. how come this s "O(n)"!?

- Anonymous February 14, 2014 | Flag
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0
of 0 vote

The Solution is O(n) -- Still can be reduced

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;

public class cubeN {
	
	public static void main(String[] args) {
		
		int sum = 0,l = 1;
		HashMap<Integer,Integer> hm  = new HashMap<Integer,Integer>();

		System.out.println("Enter the Value of N :");
		Scanner s = new Scanner(System.in);
		
		int n = s.nextInt();
							
		for(int j = 0; j < n; j++){
			
			sum = j*j*j;
			hm.put(sum,j);
		}	
				
		
		Set set = hm.entrySet();		
		Iterator i = set.iterator();
		
		
		while(i.hasNext()){
		
			Map.Entry me = (Map.Entry)i.next();
			int Value = n - (Integer) me.getKey();
			int num1 = (Integer)me.getValue();
			
			if(hm.containsKey(Value)){
				 int num2 = hm.get(Value);
				System.out.println("Combination -"+l+" "+num1+","+num2);
				l++;
				
			}		
			
		}	

	}

}

- careeradmirer January 31, 2014 | Flag Reply
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0
of 0 vote

solution in python

import math

def find_sol(a, b, N):

	# solution
	if (pow(a, 3)  + pow(b, 3) == N):
		return 1
	# over solution
	elif (pow(a, 3)  + pow(b, 3) > N):
		return 0
	# under solution
	else:
		return (find_sol(a + 1, b, N) + find_sol(a, b + 1, N)

if __name__ == '__main__':
	N = 2 # or some other value
	num_sol = find_sol(0,0,N)
	num_sol/=2

- Nino Tokuda January 31, 2014 | Flag Reply
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0
of 0 vote

Can we use binary search in this?
max value for this a^3 + b^3 will be N and mininum value will be 0.
I am sure the below code is not going to work for some cases but right now not able to figure out.
So search will be like this:

#include <math.h>
int power(int no, int power)
{
        int sum = 1;int i;
        for(i=0;i<power;i++) {
                sum = sum*no;
        }
        return sum;
}

void bsearch(int low, int high, int no) {
        printf("low %d high %d\n", low, high);
	/* base cases */
        if(low < 0 || high > no || high < 0 || low > no)
                return;
        if(power(low, 3) + power(high, 3) == no) {
                printf("%d %d\n", low, high);
                return;
        } else if(power(low, 3) + power(high, 3) >= no) {
                bsearch(low, high-1, no);
        } else
                bsearch(low+1, high, no);
}

int main()
{
        bsearch(0, 9/2, 9);
}

- aka January 31, 2014 | Flag Reply
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0
of 0 vote

initialize i =1 and j=N-1

put loop condition while(i<j)
{
if(i^3 + j^3 >N)
--j;
elseif(i^3 + j^3 <N)
++i;
else
cout<<i<<j;
++i;--j;

thats all
O(N) time.

- Nascent January 31, 2014 | Flag Reply
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0
of 0 vote

Could be done in O(N^(1/3)) (cubic root).

for (int a = 0; a * a * a <= N; ++a) {
  int b3 = N - a * a * a;
  int b = (int)pow((double)b3, 1. / 3.); // That could be done manually with binary search or something else.
  /* Here we can check close ints to be sure with precision errors */
  bool is_cubic_root = false;
  int mem = -1;
  for (int delta = -2; delta <= 2; ++delta) if (pow3(b + delta) == b3) is_cubic_root = true, mem = b + delta;
  if (is_cubic_root) {
    cout << a << ' ' << mem << endl;
  }
}

- Alex February 01, 2014 | Flag Reply
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0
of 0 vote

function findCubedPairs(n) { // assume n > 0
  var powers = {};
  var result = [];
  for (var i = 0; i < n; i++) {
    powers[i * i * i] = i;
  }
  
  for (var i = 0; i < n; i++) {
    var a = i;
    var aCubed = i*i*i;
    var bCubed = n - aCubed;
    var b;
    var neg = 1;

    if (bCubed < 0) {
      neg = -1;
    }

    var b = powers[bCubed * neg];

    if (powers[bCubed * neg]) {
      if (neg === -1) {
        result.push([a, -b]);
        result.push([-a, b]);
      } else {
        result.push([a, b]);
      }
    }
  }
  
  return result;
}

- overload119 February 02, 2014 | Flag Reply
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0
of 0 vote

void PrintAllCombination(int iNumber)
{
	//To reduce the number of iterations..... 
	int iMaxLimit = pow((double)iNumber, (double)1/3.) + 0.5 ;

	int iA, iB;
	
	int iLoopCounter = 0;
	int iFirstVal;

	//If the cube of the value is equal to number then again we are adding some value in to it, that will be waste of iteration & time.
	for( iA = 0; iA <= iMaxLimit; iA ++ )
	{
		//Calculating to reduce the multiplication process again & again.
		iFirstVal = iA * iA * iA;
		for( iB = 0; iB <= iMaxLimit; iB ++ )
		{
			if( iFirstVal + iB * iB * iB == iNumber )
				printf("A := %d B := %d\n", iA, iB);

			iLoopCounter ++;
		}
	}

	printf("Total Iteration:%d", iLoopCounter);	
}

- Kuber Saini February 12, 2014 | Flag Reply
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0
of 0 vote

Here is the optimized code that will take less time to compute the combination... please comment if I missed anything..
void PrintAllCombination(int iNumber)
{
	//To reduce the number of iterations..... 
	int iMaxLimit = pow((double)iNumber, (double)1/3.) + 0.5 ;

	int iA, iB;
	
	int iLoopCounter = 0;
	int iFirstVal;

	//If the cube of the value is equal to number then again we are adding some value in to it, that will be waste of iteration & time.
	for( iA = 0; iA <= iMaxLimit; iA ++ )
	{
		//Calculating to reduce the multiplication process again & again.
		iFirstVal = iA * iA * iA;
		for( iB = 0; iB <= iMaxLimit; iB ++ )
		{
			if( iFirstVal + iB * iB * iB == iNumber )
				printf("A := %d B := %d\n", iA, iB);

			iLoopCounter ++;
		}
	}

	printf("Total Iteration:%d", iLoopCounter);	
}

- kuber February 12, 2014 | Flag Reply
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0
of 0 vote

#include<iostream>
#include<string>
#include<math.h>
#include<stdlib.h>
using namespace std;
int main()
{
double N,t;
cin >> N;
int i;
for(i=0;i<N/2;i++)
{
t=(double)(cbrt(N-pow(i,3)));
if(t-ceil(t)==0)
{
cout<<"a is : "<<t<<" b is : " << i << endl;
}
}
}

- Xiaoyu August 19, 2014 | Flag Reply
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-2
of 2 vote

import static java.lang.System.exit;
import java.util.Scanner;

public class Number {

    int a, b;

    Number() {
        a = 0;
        b = 0;

    }

    public int[] findAB(int n) {
        int sum = 0;
        int i;
        int j;
        

            for (i = 0; i < 1000; i++) {
                for (j = 0; j < 1000; j++) {
                    sum = i * i * i + j * j * j;
                if(sum==n){
                    System.out.println(i+" , "+j);
        
                }
                }
            }
        return null;

              
    }

    public static void main(String args[]) {
        int n;
        Number num = new Number();
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        num.findAB(n);
    }

}

- Karthik Thamatam January 30, 2014 | Flag Reply
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-2
of 2 vote

import static java.lang.System.exit;
import java.util.Scanner;

public class Number {

    int a, b;

    Number() {
        a = 0;
        b = 0;

    }

    public int[] findAB(int n) {
        int sum = 0;
        int i;
        int j;
        

            for (a = 0; a < 1000; a++) {
                for (b = 0; b < 1000; b++) {
                    sum = a * a * a + b * b * b;
                if(sum==n){
                    System.out.println(a+" , "+b);
        
                }
                }
            }
        return null;

              
    }

    public static void main(String args[]) {
        int n;
        Number num = new Number();
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        num.findAB(n);
    }

}

}

- Anonymous January 30, 2014 | Flag Reply
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0
of 0 votes

#include <stdio.h>


main()
{
int a,b,i,j,k=0;
printf("enter no.");
scanf("%d",&N);

for(a=0;a<N;a++)
{
if((a*a*a)>N)
{
a--;
break;
}
k++;
}

for(i=a;i>=0;i--)
{
for(j=0;j<=a;j++)
{
if(((i*i*i)+(j*j*j))==N)
printf("%d,%d\n",i,j);
if(((i*i*i)+(j*j*j))>N)
break;
}
k++;
}
printf("\nnumber of comparison %d\n",k);
}

- dilkash.rocks May 04, 2014 | Flag


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