Check if one integer is an integer power of another

Question

This is an interview question: \"Given 2 integers x and y, check if x is an integer power of y\" (e.g. for x = 8 and y = 2 the answer is \"true\", and for x = 10 and y = 2 \

Answer 1

If you have access to the largest power of y, that can be fitted inside the required datatype, this is a really slick way of solving this problem.

Lets say, for our case, y == 3. So, we would need to check if x is a power of 3.

Given that we need to check if an integer x is a power of 3, let us start thinking about this problem in terms of what information is already at hand.

1162261467 is the largest power of 3 that can fit into an Java int.
1162261467 = 3^19 + 0

The given x can be expressed as [(a power of 3) + (some n)]. I think it is fairly elementary to be able to prove that if n is 0(which happens iff x is a power of 3), 1162261467 % x = 0.

So, to check if a given integer x is a power of three, check if x > 0 && 1162261467 % x == 0.

Generalizing. To check if a given integer x is a power of a given integer y, check if x > 0 && Y % x == 0: Y is the largest power of y that can fit into an integer datatype.

The general idea is that if A is some power of Y, A can be expressed as B/Ya, where a is some integer and A < B. It follows the exact same principle for A > B. The A = B case is elementary.

Answer 2

in the case the number is too large ... use log function to reduce time complexity:

import math
base = int(input("Enter the base number: "))
for i in range(base,int(input("Enter the end of range: "))+1):
    if(math.log(i) / math.log(base) % 1 == 0 ):
        print(i)

Answer 3

Here is a Python version which puts together the ideas of @salva and @Axn and is modified to not generate any numbers greater than those given and uses only simple storage (read, "no lists") by repeatedly paring away at the number of interest:

def perfect_base(b, n):
    """Returns True if integer n can be expressed as b**e where
    n is a positive integer, else False."""

    assert b > 1 and n >= b and int(n) == n and int(b) == b

    # parity check
    if not b % 2:
        if n % 2:
            return False  # b,n is even,odd
        if b == 2:
            return n & (n - 1) == 0
        if not b & (b - 1) and n & (n - 1):
            return False  # b == 2**m but n != 2**M
    elif not n % 2:
        return False  # b,n is odd,even

    while n >= b:
        d = b
        while d <= n:
            n, r = divmod(n, d)
            if r:
                return False
            d *= d
    return n == 1

Answer 4

I would implement the function like so:

bool IsWholeNumberPower(int x, int y)
{
    double power = log(x)/log(y);
    return floor(power) == power;
}

This shouldn't need check within a delta as is common with floating point comparisons, since we're checking whole numbers.

Answer 5

This looks for the exponent in O(log N) steps:

#define MAX_POWERS 100

int is_power(unsigned long x, unsigned long y) {
  int i;
  unsigned long powers[MAX_POWERS];
  unsigned long last;
  last = powers[0] = y;

  for (i = 1; last < x; i++) {
    last *= last; // note that last * last can overflow here!
    powers[i] = last;
  }
  while (x >= y) {
    unsigned long top = powers[--i];
    if (x >= top) {
      unsigned long x1 = x / top;
      if (x1 * top != x) return 0;
      x = x1;
    }
  }
  return (x == 1);
}

Negative numbers are not handled by this code, but it can be done easyly with some conditional code when i = 1

Answer 6

This looks to be pretty fast for positive numbers as it finds the lower and upper limits for desired power and then applies binary search.

#include <iostream>
#include <cmath>
using namespace std;

//x is the dividend, y the divisor.
bool isIntegerPower(int x, int y)
{
    int low = 0, high;
    int exp = 1;
    int val = y;
    //Loop by changing exponent in the powers of 2 and
    //Find out low and high exponents between which the required exponent lies.
    while(1)
    {
        val = pow((double)y, exp);
        if(val == x)
            return true;
        else if(val > x)
            break;
        low = exp;
        exp = exp * 2;
        high = exp;
    }
    //Use binary search to find out the actual integer exponent if exists
    //Otherwise, return false as no integer power.
    int mid = (low + high)/2;
    while(low < high)
    {
        val = pow((double)y, mid);
        if(val > x)
        {
            high = mid-1;
        }
        else if(val == x)
        {
            return true;
        }
        else if(val < x)
        {
            low = mid+1;
        }
        mid = (low + high)/2;
    }
    return false;
}

int main()
{
    cout<<isIntegerPower(1024,2);
}