Tricky Google interview question

Question

A friend of mine is interviewing for a job. One of the interview questions got me thinking, just wanted some feedback.

There are 2 non-negative integers: i and j. Given the following equation, find an (optimal) solution to iterate over i and j in such a way that the output is sorted.

2^i * 5^j

So the first few rounds would look like this:

2^0 * 5^0 = 1
2^1 * 5^0 = 2
2^2 * 5^0 = 4
2^0 * 5^1 = 5
2^3 * 5^0 = 8
2^1 * 5^1 = 10
2^4 * 5^0 = 16
2^2 * 5^1 = 20
2^0 * 5^2 = 25

Try as I might, I can't see a pattern. Your thoughts?

Answer 1

Dijkstra derives an eloquent solution in "A Discipline of Programming". He attributes the problem to Hamming. Here is my implementation of Dijkstra’s solution.

int main()
{
    const int n = 20;       // Generate the first n numbers

    std::vector<int> v(n);
    v[0] = 1;

    int i2 = 0;             // Index for 2
    int i5 = 0;             // Index for 5

    int x2 = 2 * v[i2];     // Next two candidates
    int x5 = 5 * v[i5];

    for (int i = 1; i != n; ++i)
    {
        int m = std::min(x2, x5);
        std::cout << m << " ";
        v[i] = m;

        if (x2 == m)
        {
            ++i2;
            x2 = 2 * v[i2];
        }
        if (x5 == m)
        {
            ++i5;
            x5 = 5 * v[i5];
        }
    }

    std::cout << std::endl;
    return 0;
}

Answer 2

If you go by what's really happening when we increment i or j in the expression 2^i * 5^j, you are either multiplying by another 2 or another 5. If we restate the problem as - given a particular value of i and j, how would you find the next greater value, the solution becomes apparent.

Here are the rules we can quite intuitively enumerate:

• If there is a pair of 2s (i > 1) in the expression, we should replace them with a 5 to get the next biggest number. Thus, i -= 2 and j += 1.
• Otherwise, if there is a 5 (j > 0), we need to replace it with three 2s. So j -= 1 and i += 3.
• Otherwise, we need to just supply another 2 to increase the value by a minimum. i += 1.

Here's the program in Ruby:

i = j = 0                                                                       
20.times do                                                                     
  puts 2**i * 5**j

  if i > 1                                                                      
    j += 1                                                                      
    i -= 2                                                                      
  elsif j > 0                                                                   
    j -= 1                                                                      
    i += 3                                                                      
  else                                                                          
    i += 1                                                                      
  end                                                                                                                                                               
end

Answer 3

I know I am likely wrong but there is a very simple heuristic here since it does not involve many numbers like 2,3,5. We know that for any i,j 2^i * 5^j next sequence would be 2^(i-2) * 5^(j+1). Being a google q it must have a simple solution.

def func(i, j):
 print i, j, (2**i)*(5**j)

imax=i=2
j=0
print "i", "j", "(2**i)*(5**j)"

for k in range(20):
    func(i,j)
    j=j+1; i=i-2
    if(i<0):
        i = imax = imax+1
        j=0

This produces output as :

i j (2**i)*(5**j)
2 0 4
0 1 5
3 0 8
1 1 10
4 0 16
2 1 20
0 2 25
5 0 32
3 1 40
1 2 50
6 0 64
4 1 80
2 2 100
0 3 125
7 0 128
5 1 160
3 2 200
1 3 250
8 0 256
6 1 320

Answer 4

Just was curious what to expect next week and have found this question.

I think, the idea is 2^i increases not in that big steps as 5^j. So increase i as long as next j-step wouldn't be bigger.

The example in C++ (Qt is optional):

QFile f("out.txt"); //use output method of your choice here
f.open(QIODevice::WriteOnly);
QTextStream ts(&f);

int i=0;
int res=0;
for( int j=0; j<10; ++j )
{
    int powI = std::pow(2.0,i );
    int powJ = std::pow(5.0,j );
    while ( powI <= powJ  ) 
    {
        res = powI * powJ;
        if ( res<0 ) 
            break; //integer range overflow

        ts<<i<<"\t"<<j<<"\t"<<res<<"\n";
        ++i;
        powI = std::pow(2.0,i );

    }
}

The output:

i   j   2^i * 5^j
0   0   1
1   1   10
2   1   20
3   2   200
4   2   400
5   3   4000
6   3   8000
7   4   80000
8   4   160000
9   4   320000
10  5   3200000
11  5   6400000
12  6   64000000
13  6   128000000
14  7   1280000000

Answer 5

Here is my solution

#include <stdio.h>
#include <math.h>
#define N_VALUE 5
#define M_VALUE  5

int n_val_at_m_level[M_VALUE];

int print_lower_level_val(long double val_of_higher_level, int m_level)
{
int  n;
long double my_val;


for( n = n_val_at_m_level[m_level]; n <= N_VALUE; n++) {
    my_val =  powl(2,n) * powl(5,m_level);
    if(m_level != M_VALUE && my_val > val_of_higher_level) {
        n_val_at_m_level[m_level] = n;
        return 0;
    }
    if( m_level != 0) {
        print_lower_level_val(my_val, m_level - 1);
    }
    if(my_val < val_of_higher_level || m_level == M_VALUE) {
        printf("    %Lf n=%d m = %d\n", my_val, n, m_level);
    } else {
        n_val_at_m_level[m_level] = n;
        return 0;
    }
 }
 n_val_at_m_level[m_level] = n;
 return 0;
 }


 main()
 {
    print_lower_level_val(0, M_VALUE); /* to sort 2^n * 5^m */
 }

Result :

1.000000 n = 0 m = 0
2.000000 n = 1 m = 0
4.000000 n = 2 m = 0
5.000000 n = 0 m = 1
8.000000 n = 3 m = 0
10.000000 n = 1 m = 1
16.000000 n = 4 m = 0
20.000000 n = 2 m = 1
25.000000 n = 0 m = 2
32.000000 n = 5 m = 0
40.000000 n = 3 m = 1
50.000000 n = 1 m = 2
80.000000 n = 4 m = 1
100.000000 n = 2 m = 2
125.000000 n = 0 m = 3
160.000000 n = 5 m = 1
200.000000 n = 3 m = 2
250.000000 n = 1 m = 3
400.000000 n = 4 m = 2
500.000000 n = 2 m = 3
625.000000 n = 0 m = 4
800.000000 n = 5 m = 2
1000.000000 n = 3 m = 3
1250.000000 n = 1 m = 4
2000.000000 n = 4 m = 3
2500.000000 n = 2 m = 4
3125.000000 n = 0 m = 5
4000.000000 n = 5 m = 3
5000.000000 n = 3 m = 4
6250.000000 n = 1 m = 5
10000.000000 n = 4 m = 4
12500.000000 n = 2 m = 5
20000.000000 n = 5 m = 4
25000.000000 n = 3 m = 5
50000.000000 n = 4 m = 5
100000.000000 n = 5 m = 5

Answer 6

If you draw a matrix with i as the row and j as the column you can see the pattern. Start with i = 0 and then just traverse the matrix by going up 2 rows and right 1 column until you reach the top of the matrix (j >= 0). Then go i + 1, etc...

So for i = 7 you travel like this:

7, 0 -> 5, 1 -> 3, 2 -> 1, 3

And for i = 8:

8, 0 -> 6, 1 -> 4, 2 -> 2, 3 -> 0, 4

Here it is in Java going up to i = 9. It prints the matrix position (i, j) and the value.

for(int k = 0; k < 10; k++) {

    int j = 0;

    for(int i = k; i >= 0; i -= 2) {

        int value = (int)(Math.pow(2, i) * Math.pow(5, j));
        System.out.println(i + ", " + j + " -> " + value);
        j++;
    }
}

Answer 7

Use a Min-heap.

Put 1.

extract-Min. Say you get x.

Push 2x and 5x into the heap.

Repeat.

Instead of storing x = 2^i * 5^j, you can store (i,j) and use a custom compare function.

Answer 8

You have to keep track of the individual exponents of them, and what their sums would be

so you start with f(0,0) --> 1 now you have to increment one of them:

f(1,0) = 2
f(0,1) = 5

so we know 2 is the next - we also know we can increment i's exponent up until the sum surpases 5.

You keep going back and forth like this until you're at your deisred number of rounds.

Answer 9

calculate the results and put them in a sorted list, together with the values for i and j

Answer 10

Using dynamic programming you can do this in O(n). Ground truth is that no values of i and j can give us 0, and to get 1 both values must be 0;

TwoCount[1] = 0
FiveCount[1] = 0

// function returns two values i, and j
FindIJ(x) {
    if (TwoCount[x / 2]) {
        i = TwoCount[x / 2] + 1
        j = FiveCount[x / 2]
    }
    else if (FiveCount[x / 5]) {
        i = TwoCount[x / 2]
        j = FiveCount[x / 5] + 1
    }
}

Whenever you call this function check if i and j are set, if they are not null, then populate TwoCount and FiveCount

---

C++ answer. Sorry for bad coding style, but i'm in a hurry :(

#include <cstdlib>
#include <iostream>
#include <vector>

int * TwoCount;
int * FiveCount;

using namespace std;

void FindIJ(int x, int &i, int &j) {
        if (x % 2 == 0 && TwoCount[x / 2] > -1) {
                cout << "There's a solution for " << (x/2) << endl;
                i = TwoCount[x / 2] + 1;
                j = FiveCount[x / 2];
        } else if (x % 5 == 0 && TwoCount[x / 5] > -1) {
                cout << "There's a solution for " << (x/5) << endl;
                i = TwoCount[x / 5];
                j = FiveCount[x / 5] + 1;
        }    
}

int main() {
        TwoCount = new int[200];
        FiveCount = new int[200];

        for (int i = 0; i < 200; ++i) {
                TwoCount[i] = -1;
                FiveCount[i] = -1;
        }

        TwoCount[1] = 0;
        FiveCount[1] = 0;

        for (int output = 2; output < 100; output++) {
                int i = -1;
                int j = -1;
                FindIJ(output, i, j);
                if (i > -1 && j > -1) {
                        cout << "2^" << i << " * " << "5^" 
                                     << j << " = " << output << endl;
                        TwoCount[output] = i;
                        FiveCount[output] = j;
                }
        }    
}

Obviously you can use data structures other than array to dynamically increase your storage etc. This is just a sketch to prove that it works.

Answer 11

You know that log_2(5)=2.32. From this we note that 2^2 < 5 and 2^3 > 5.

Now look a matrix of possible answers:

j/i  0   1   2   3   4   5
 0   1   2   4   8  16  32
 1   5  10  20  40  80 160 
 2  25  50 100 200 400 800
 3 125 250 500 ...

Now, for this example, choose the numbers in order. There ordering would be:

j/i  0   1   2   3   4   5
 0   1   2   3   5   7  10
 1   4   6   8  11  14  18
 2   9  12  15  19  23  27
 3  16  20  24...

Note that every row starts 2 columns behind the row starting it. For instance, i=0 j=1 comes directly after i=2 j=0.

An algorithm we can derive from this pattern is therefore (assume j>i):

int i = 2;
int j = 5;
int k;
int m;

int space = (int)(log((float)j)/log((float)i));
for(k = 0; k < space*10; k++)
{
    for(m = 0; m < 10; m++)
    {
        int newi = k-space*m;
        if(newi < 0)
            break;
        else if(newi > 10)
            continue;
        int result = pow((float)i,newi) * pow((float)j,m);
        printf("%d^%d * %d^%d = %d\n", i, newi, j, m, result);
    }
}   

NOTE: The code here caps the values of the exponents of i and j to be less than 10. You could easily extend this algorithm to fit into any other arbitrary bounds.

NOTE: The running time for this algorithm is O(n) for the first n answers.

NOTE: The space complexity for this algorithm is O(1)

Answer 12

This is very easy to do O(n) in functional languages. The list l of 2^i*5^j numbers can be simply defined as 1 and then 2*l and 5*l merged. Here is how it looks in Haskell:

merge :: [Integer] -> [Integer] -> [Integer]
merge (a:as) (b:bs)   
  | a < b   = a : (merge as (b:bs))
  | a == b  = a : (merge as bs)
  | b > a   = b : (merge (a:as) bs)

xs :: [Integer]
xs = 1 : merge (map(2*)xs) (map(5*)xs)

The merge function gives you a new value in constant time. So does map and hence so does l.

Answer 13

A FIFO-based solution needs less storage capacity. Python code.

F = [[1, 0, 0]]             # FIFO [value, i, j]
i2 = -1; n2 = n5 = None     # indices, nexts
for i in range(1000):       # print the first 1000
    last = F[-1][:]
    print "%3d. %21d = 2^%d * 5^%d" % tuple([i] + last)
    if n2 <= last: i2 += 1; n2 = F[i2][:]; n2[0] *= 2; n2[1] += 1
    if n5 <= last: i2 -= 1; n5 = F.pop(0); n5[0] *= 5; n5[2] += 1
    F.append(min(n2, n5))

output:

  0.                     1 = 2^0 * 5^0
  1.                     2 = 2^1 * 5^0
  2.                     4 = 2^2 * 5^0
 ...
998. 100000000000000000000 = 2^20 * 5^20
999. 102400000000000000000 = 2^27 * 5^17

Answer 14

The algorithm implemented by user515430 by Edsger Dijkstra (http://www.cs.utexas.edu/users/EWD/ewd07xx/EWD792.PDF) is probably as fast as you can get. I call every number that is a form of 2^i * 5^j a "special number". Now vlads answer would be O(i*j) but with a double algorithm, one to generate the special numbers O(i*j) and one to sort them (according to the linked article also O(i*j).

But let's check Dijkstra's algorithm (see below). In this case n is the amount of special numbers we are generating, so equal to i*j. We are looping once, 1 -> n and in every loop we perform a constant action. So this algorithm is also O(i*j). And with a pretty blazing fast constant too.

My implementation in C++ with GMP (C++ wrapper), and dependancy on boost::lexical_cast, though that can be easily remove (I'm lazy, and who doesn't use Boost?). Compiled with g++ -O3 test.cpp -lgmpxx -o test. On Q6600 Ubuntu 10.10 time ./test 1000000 gives 1145ms.

#include <iostream>
#include <boost/lexical_cast.hpp>
#include <gmpxx.h>

int main(int argc, char *argv[]) {
    mpz_class m, x2, x5, *array, r;
    long n, i, i2, i5;

    if (argc < 2) return 1;

    n = boost::lexical_cast<long>(argv[1]);

    array = new mpz_class[n];
    array[0] = 1;

    x2 = 2;
    x5 = 5;
    i2 = i5 = 0;

    for (i = 1; i != n; ++i) {
        m = std::min(x2, x5);

        array[i] = m;

        if (x2 == m) {
            ++i2;
            x2 = 2 * array[i2];
        }

        if (x5 == m) {
            ++i5;
            x5 = 5 * array[i5];
        }
    }

    delete [] array;
    std::cout << m << std::endl;

    return 0;
}

Answer 15

here is a more refined way of doing it (more refined than my previous answer, that is):

imagine the numbers are placed in a matrix:

     0    1    2    3    4    5   -- this is i
----------------------------------------------
0|   1    2    4    8   16   32
1|   5   10   20   40   80  160
2|  25   50  100  200  400  800
3| 125  250  500 1000 2000 ...
4| 625 1250 2500 5000 ...
j on the vertical

what you need to do is 'walk' this matrix, starting at (0,0). You also need to keep track of what your possible next moves are. When you start at (0,0) you only have two options: either (0,1) or (1,0): since the value of (0,1) is smaller, you choose that. then do the same for your next choice (0,2) or (1,0). So far, you have the following list: 1, 2, 4. Your next move is (1,0) since the value there is smaller than (0,3). However, you now have three choices for your next move: either (0,3), or (1,1), or (2,0).

You don't need the matrix to get the list, but you do need to keep track of all your choices (i.e. when you get to 125+, you will have 4 choices).

Answer 16

Why not try looking at this from the other direction. Use a counter to test the possible answers against the original formula. Sorry for the pseudo code.

for x = 1 to n
{
  i=j=0
  y=x
  while ( y > 1 )
  {
    z=y
    if y divisible by 2 then increment i and divide y by 2
    if y divisible by 5 then increment j and divide y by 5

    if y=1 then print i,j & x  // done calculating for this x

    if z=y then exit while loop  // didn't divide anything this loop and this x is no good 
  }
}

Answer 17

This is the relevant entry at OEIS.

It seems to be possible to obtain the ordered sequence by generating the first few terms, say

1 2 4 5

and then, starting from the second term, multiplying by 4 and 5 to get the next two

1 2 4 5 8 10

1 2 4 5 8 10 16 20

1 2 4 5 8 10 16 20 25

and so on...

Intuitively, this seems correct, but of course a proof is missing.

Answer 18

Here is my attempt with Scala:

case class IndexValue(twosIndex: Int, fivesIndex: Int)
case class OutputValues(twos: Int, fives: Int, value: Int) {
  def test(): Boolean = {
    Math.pow(2,  twos) * Math.pow(5, fives) == value
  }
}

def run(last: IndexValue = IndexValue(0, 0), list: List[OutputValues] = List(OutputValues(0, 0, 1))): List[OutputValues] = {
  if (list.size > 20) {
    return list
  }

  val twosValue = list(last.twosIndex).value * 2
  val fivesValue = list(last.fivesIndex).value * 5

  if (twosValue == fivesValue) {
    val lastIndex = IndexValue(last.twosIndex + 1, last.fivesIndex + 1)
    val outputValues = OutputValues(value = twosValue, twos = list(last.twosIndex).twos + 1, fives = list(last.fivesIndex).fives + 1)
    run(lastIndex, list :+ outputValues)
  } else if (twosValue < fivesValue) {
    val lastIndex = IndexValue(last.twosIndex + 1, last.fivesIndex)
    val outputValues = OutputValues(value = twosValue, twos = list(last.twosIndex).twos + 1, fives = list(last.twosIndex).fives)
    run(lastIndex, list :+ outputValues)
  } else {
    val lastIndex = IndexValue(last.twosIndex, last.fivesIndex + 1)
    val outputValues = OutputValues(value = fivesValue, twos = list(last.fivesIndex).twos, fives = list(last.fivesIndex).fives + 1)
    run(lastIndex, list :+ outputValues)
  }
}

val initialIndex = IndexValue(0, 0)
run(initialIndex, List(OutputValues(0, 0, 1))) foreach println

Output:

OutputValues(0,0,1)
OutputValues(1,0,2)
OutputValues(2,0,4)
OutputValues(0,1,5)
OutputValues(3,0,8)
OutputValues(1,1,10)
OutputValues(4,0,16)
OutputValues(2,1,20)
OutputValues(0,2,25)
OutputValues(5,0,32)
OutputValues(3,1,40)
OutputValues(1,2,50)
OutputValues(6,0,64)
OutputValues(4,1,80)
OutputValues(2,2,100)
OutputValues(0,3,125)
OutputValues(7,0,128)
OutputValues(5,1,160)
OutputValues(3,2,200)
OutputValues(1,3,250)
OutputValues(8,0,256)

Answer 19

If we are allowed to use java Collection then we can have these number in O(n^2)

public static void main(String[] args) throws Exception {
    int powerLimit = 7;  
     int first = 2;
     int second = 5;
    SortedSet<Integer> set = new TreeSet<Integer>();

    for (int i = 0; i < powerLimit; i++) {
        for (int j = 0; j < powerLimit; j++) {
            Integer x = (int) (Math.pow(first, i) * Math.pow(second, j));
            set.add(x);
        }
    }

    set=set.headSet((int)Math.pow(first, powerLimit));

    for (int p : set)
        System.out.println(p);
}

Here powerLimit has to be initialised very carefully !! Depending upon how many numbers you want.

Answer 20

My Intuition :

If I take initial value as 1 where i=0, j=0, then I can create next numbers as (2^1)(5^0), (2^2)(5^0), (2^0)*(5^1), ... i.e 2,4,5 ..

Let say at any point my number is x. then I can create next numbers in the following ways :

• x * 2
• x * 4
• x * 5

Explanation :

Since new numbers can only be the product with 2 or 5.
But 4 (pow(2,2)) is smaller than 5, and also we have to generate 
Numbers in sorted order.Therefore we will consider next numbers
be multiplied with 2,4,5.
Why we have taken x*4 ? Reason is to pace up i, such that it should not 
be greater than pace of j(which is 5 to power). It means I will 
multiply my number by 2, then by 4(since 4 < 5), and then by 5 
to get the next three numbers in sorted order.

Test Run

We need to take an Array-list of Integers, let say Arr.

Also put our elements in Array List<Integers> Arr.
Initially it contains Arr : [1]

• Lets start with x = 1.

  Next three numbers are 1*2, 1*4, 1*5 [2,4,5]; Arr[1,2,4,5]

• Now x = 2

  Next three numbers are [4,8,10] {Since 4 already occurred we will ignore it} [8,10]; Arr[1,2,4,5,8,10]

• Now x =4

  Next three numbers [8,16,20] {8 already occurred ignore it} [16,20] Arr[1,2,4,5,8,10,16,20]

• x = 5

  Next three numbers [10,20,25] {10,20} already so [25] is added Arr[1,2,4,5,8,10,16,20,25]

Termination Condition

 Terminating condition when Arr last number becomes greater 
 than (5^m1 * 2^m2), where m1,m2 are given by user.

Analysis

 Time Complexity : O(K) : where k is numbers possible between i,j=0 to 
 i=m1,j=m2.
 Space Complexity : O(K)

Answer 21

My implementation is based on the following ideas:

• Use two queues Q2 and Q5, both initialized with 1. We will keep both queue in sorted order.
• At every step, dequeue the smallest number element MIN from Q2 or Q5 and print it. If both Q2 and Q5 have the same element - remove both. Print this number. This is basically merging of two sorted arrays - at each step choose the smallest element and advance.
• Enqueue MIN*2 to Q2 and MIN*5 to Q5. This change does not break the invariant of Q2/Q5 being sorted, because MIN is higher than previous MIN number.

Example:

Start with 1 and 1 (to handle i=0;j=0 case):
  Q2: 1
  Q5: 1
Dequeue 1, print it and enqueue 1*2 and 1*5:
  Q2: 2
  Q5: 5
Pick 2 and add 2*2 and 2*5:
  Q2: 4
  Q5: 5 10
Pick 4 and add 4*2 and 4*5:
  Q2: 8
  Q5: 5 10 20
....

Code in Java:

public void printNumbers(int n) {
    Queue<Integer> q2 = new LinkedList<Integer>();
    Queue<Integer> q5 = new LinkedList<Integer>();
    q2.add(1);
    q5.add(1);
    for (int i = 0; i < n; i++) {
        int a = q2.peek();
        int b = q5.peek();
        int min = Math.min(a, b);
        System.out.println(min);
        if (min == a) {
            q2.remove();
        }
        if (min == b) {
            q5.remove();
        }
        q2.add(min * 2);
        q5.add(min * 5);
    }
}