Given an integer, write a function to determine if it is a power of three.
Follow up:
Could you do it without using any loop / recursion?
Credits:
Special thanks to
@dietpepsi for adding this problem and creating all test cases.
Brute-force Solution:
The problem asks for if there is a number n, where n = 3^x. So the brute-force solution is to try each x from 0 and see if 3^x is equal to n, until 3^x is greater than n then we return false.
Code (Java):
public class Solution {
public boolean isPowerOfThree(int n) {
if (n <= 0) {
return false;
}
int i = 0;
long cur = (long) Math.pow(3, i);
long longN = (long) n;
while (cur <= longN) {
if (cur == longN) {
return true;
}
i++;
cur = (long) Math.pow(3, i);
}
return false;
}
}
Some tricky solutions:
Method 2
If
log10(n) / log10(3) returns an int (more precisely, a double but has 0 after decimal point), then n is a power of 3. (
original post). But
be careful here, you cannot use
log (natural log) here, because it will generate round off error for
n=243. This is more like a coincidence. I mean when
n=243, we have the following results:
log(243) = 5.493061443340548 log(3) = 1.0986122886681098
==> log(243)/log(3) = 4.999999999999999
log10(243) = 2.385606273598312 log10(3) = 0.47712125471966244
==> log10(243)/log10(3) = 5.0
This happens because log(3) is actually slightly larger than its true value due to round off, which makes the ratio smaller.
public boolean isPowerOfThree(int n) {
return (Math.log10(n) / Math.log10(3)) % 1 == 0;
}
Editional Soluton:
Solution
In this article we will look into ways of speeding up simple computations and why that is useful in practice.
Approach #1 Loop Iteration [Accepted]
One simple way of finding out if a number n is a power of a number b is to keep dividing n by b as long as the remainder is 0. This is because we can write
n=bxn=b×b×…×b
Hence it should be possible to divide n by b x times, every time with a remainder of 0 and the end result to be 1.
Java
public class Solution {
public boolean isPowerOfThree(int n) {
if (n < 1) {
return false;
}
while (n % 3 == 0) {
n /= 3;
}
return n == 1;
}
}
Notice that we need a guard to check that n != 0, otherwise the while loop will never finish. For negative numbers, the algorithm does not make sense, so we will include this guard as well.
Complexity Analysis
Time complexity :
O(logb(n)). In our case that is
O(log3n). The number of divisions is given by that logarithm.
Space complexity :
O(1). We are not using any additional memory.
Approach #2 Base Conversion [Accepted]
In Base 10, all powers of 10 start with the digit
1 and then are followed only by
0 (e.g. 10, 100, 1000). This is true for other bases and their respective powers. For instance in
base 2, the representations of
102,
1002 and
10002 are
210,
410 and
810 respectively. Therefore if we convert our number to base 3 and the representation is of the form 100...0, then the number is a power of 3.
Proof
Given the base 3 representation of a number as the array s, with the least significant digit on index 0, the formula for converting from base 3 to base 10 is:
i=0∑len(s)−1s[i]∗3i
Therefore, having just one digit of 1 and everything else 0 means the number is a power of 3.
Implementation
All we need to do is convert [4] the number to base 3 and check if it is written as a leading 1 followed by all 0.
A couple of built-in Java functions will help us along the way.
String baseChange = Integer.toString(number, base);
The code above converts number into base base and returns the result as a String. For example, Integer.toString(5, 2) == "101" andInteger.toString(5, 3) == "12".
boolean matches = myString.matches("123");
The code above checks if a certain Regular Expression[2] pattern exists inside a string. For instance the above will return true if the substring "123" exists inside the string myString.
boolean powerOfThree = baseChange.matches("^10*$")
We will use the regular expression above for checking if the string starts with 1 ^1, is followed by zero or more 0s 0* and contains nothing else $.
Java
public class Solution {
public boolean isPowerOfThree(int n) {
return Integer.toString(n, 3).matches("^10*$");
}
}
Complexity Analysis
Approach #3 Mathematics [Accepted]
We can use mathematics as follows
n=3ii=log3(n)i=logb(n)logb(3)
n is a power of three if and only if i is an integer. In Java, we check if a number is an integer by taking the decimal part (using % 1) and checking if it is 0.
Java
public class Solution {
public boolean isPowerOfThree(int n) {
return (Math.log10(n) / Math.log10(3)) % 1 == 0;
}
}
Common pitfalls
This solution is problematic because we start using doubles, which means we are subject to precision errors. This means, we should never use == when comparing doubles. That is because the result of Math.log10(n) / Math.log10(3) could be 5.0000001 or 4.9999999. This effect can be observed by using the function Math.log() instead of Math.log10().
In order to fix that, we need to compare the result against an epsilon.
Java
return (Math.log(n) / Math.log(3) + epsilon) % 1 <= 2 * epsilon;
Complexity Analysis
Time complexity :
Unknown The expensive operation here is
Math.log, which upper bounds the time complexity of our algorithm. The implementation is dependent on the language we are using and the compiler
[3]
Space complexity :
O(1). We are not using any additional memory. The
epsilon variable can be inlined.
Approach #4 Integer Limitations [Accepted]
An important piece of information can be deduced from the function signature
public boolean isPowerOfThree(int n)
In particular, n is of type int. In Java, this means it is a 4 byte, signed integer [ref]. The maximum value of this data type is 2147483647. Three ways of calculating this value are
- Google
System.out.println(Integer.MAX_VALUE);- MaxInt = 2232−1 since we use 32 bits to represent the number, half of the range is used for negative numbers and 0 is part of the positive numbers
Knowing the limitation of n, we can now deduce that the maximum value of n that is also a power of three is 1162261467. We calculate this as:
3⌊log3MaxInt⌋=3⌊19.56⌋=319=1162261467
Therefore, the possible values of
n where we should return
true are
30,
31 ...
319. Since 3 is a prime number, the only divisors of
319 are
30,
31 ...
319, therefore all we need to do is divide
319 by
n. A remainder of
0 means
n is a divisor of
319 and therefore a power of three.
Java
public class Solution {
public boolean isPowerOfThree(int n) {
return n > 0 && 1162261467 % n == 0;
}
}
Complexity Analysis
Time complexity :
O(1). We are only doing one operation.
Space complexity :
O(1). We are not using any additional memory.
Single runs of the function make it is hard to accurately measure the difference of the two solutions. On LeetCode, on the Accepted Solutions Runtime Distribution page, all solutions being between 15 ms and 20 ms. For completeness, we have proposed the following benchmark to see how the two solutions differ.
Java Benchmark Code
public static void main(String[] args) {
Solution sol = new Solution();
int iterations = 1; // See table header for this value
for (int i = 0; i < iterations; i++) {
sol.isPowerOfThree(i);
}
}
In the table below, the values are in seconds.
| Iterations | 106 | 107 | 108 | 109 | Maxint |
|---|
| Java Approach #1 (Naive) | 0.04 | 0.07 | 0.30 | 2.47 | 5.26 |
| Java Approach #2 (Strings) | 0.68 | 4.02 | 38.90 | 409.16 | 893.89 |
| Java Approach #3 (Logarithms) | 0.09 | 0.50 | 4.59 | 45.53 | 97.50 |
| Java Approach #4 (Fast) | 0.04 | 0.06 | 0.08 | 0.41 | 0.78 |
As we can see, for small values of N, the difference is not noticeable, but as we do more iterations and the values of n passed to isPowerOfThree() grow, we see significant boosts in performance for Approach #4.
Conclusion
Simple optimizations like this might seem negligible, but historically, when computation power was an issue, it allowed certain computer programs (such as Quake 3[1]) possible.
References
