If the n==k and k==1, given number (n) is the power of k. Return true. To solve the problem follow the below idea: A simple method for this is to simply take the log of the number on base 2 and if you get an integer then the number is the power of 2 Below is the implementation of the above approach: C++ C I used the cross join to make sure I get enough rows, and sys.columns because I know it has at least 40-50 rows. Submitted by Bipin Kumar, on November 21, 2019. The good thing is, that you can extend this to powers of 4, 5, 6, 7, whatever. 1/n! @Graipher How about tail-call optimization? Finally, it needs to check whether the result is 1, 3, 9, 27, 81, or 243, which is done by some simple hashing that I found by trial-and-error. STEP 1: Initially, we will import the math module in the program. It is inspired by the solution to check if a number is power of 2 by doing n & (n - 1) == 0, Time complexity: O(log(n)) depending on language and compiler, space complexity: O(log(n)), If n = 3^i, then i = log(n) / log(3), and thus comes to the solution, Time complexity: depending on language and compiler, space complexity: O(1), Because 3^19 = 1162261467 is the largest power of 3 number fits in a 32 bit integer, thus we can do, Time complexity: O(1), space complexity: O(1). Does a knockout punch always carry the risk of killing the receiver? True in mathematical sense, not practical because of rounding errors. 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The best answers are voted up and rise to the top, Not the answer you're looking for? +1. Let q denote the number of pairings of the form (1 0) and let r denote the number of pairings of the form (0 1). Complexity of your code is very, very bad if n == 0 or k == 1 because your loop doesn't finish :-( So that should be fixed. just for information. This sacrifices some speed in the case that the number is a power of k, but gains about a factor 2 if it is not: And, finally, some timings. Now, we will write the Python program by the implementation of the above algorithm. As long as you have multiplication in hardware, % is going to be faster. Returns the symmetric difference between two lists, after applying the provided function to each list element of both: We are closing our Disqus commenting system for some maintenanace issues. Required fields are marked *. General Solution to check if any number N is a power of P: Refer to the code below for implementation. Storing there an arbitrary power of three works. 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Here are Saturdays winning lottery numbers: 15-45-64-67-68, Powerball: 18, Power Play: 2X. Have a look at the timings I took on my machine for a comparison of yours and my (fixed) algorithm in my updated answer. And, no, I'm not inclined to test this ! That's correct, but my trivial initialization like. Manage Settings Yes. We and our partners use data for Personalised ads and content, ad and content measurement, audience insights and product development. Why not doing it the (straight) forward way? For example, if a given number ends with 1, it must be a 4th or 8th or 12th and so on power of three, if at all. This is pseudo code, it can easily be implemented in C or C++ efficiently. Can I drink black tea thats 13 years past its best by date? "power of N?" Could it be that Python optimizes the call into "jump"? @AlexandreSantos exactly! Your comments are very useful, thanks. The fastest solution is either testing if n > 0 && 3**19 % n == 0 as given in another answer or perfect hashing (below). and Twitter for latest update. Let your candidate value be T, initialized to 1. [CDATA[ I divide by 10^3 and get a quotient of 0 and a remainderof 100. Any integer number other than power of 3 which divides highest power of 3 value that integer can hold 3^19 = 1162261467 (Assuming that integers are stored using 32 bits) will give reminder non-zero. Essentially you do a binary search on powers of 3. % uses the euclidian algorithm which is a general algorithm to determine the remainder when dviding by an arbitriary number. Is abiogenesis virtually impossible from a probabilistic standpoint without a multiverse? Check if the number is divisible by 3, if yes then keep checking the same for number/3 recursively. Check if one integer is an integer power of another, Determining the next highest number that can be factored into prime numbers {2,3,5,7}. The example implementation shown below is written in Python but should not be difficult to rewrite in other languages if needed. Constraints: -100000 <= N <= 100000. A $2 ticket gives you a one in 292.2 million chance at joining the hall of Powerball jackpot champions. O(1). @greybeard. +1 for the correct approach, but recursion (in its true sense) is completely unnecessary. I share tutorials of PHP, Python, Javascript, JQuery, Laravel, Livewire, Codeigniter, Node JS, Express JS, Vue JS, Angular JS, React Js, MySQL, MongoDB, REST APIs, Windows, Xampp, Linux, Ubuntu, Amazon AWS, Composer, SEO, WordPress, SSL and Bootstrap from a starting stage. There are several ways to see if a number is a perfect cube. I will comment here, since it is currently (and I hope remains) the top-voted answer. There exists a constant time (pretty fast) method for integers of limited size (e.g. Write a Python program to check if a given positive integer is a power of three. Use of and/or registration on any portion of this site constitutes acceptance of our User Agreement (updated 4/4/2023), Privacy Policy and Cookie Statement, and Your Privacy Choices and Rights (updated 1/26/2023). check if number is power of 2 python Comment . So when the result of an instrumented function is used, and it's not about getting to know it, pay attention to the fact that the compiler might optimize out unknown parts, take a look at the disassembly. Next: Write a Python program to check if a given positive integer is a power of four. With SET STATISTICS TIME ON it record the lowest possible, 1 millisecond. This generates the values on compile-time. Another approach is to generate a table on compile time. Yes. Python Program to Find Sum of Series 1/1! To do binary search, first you need a fast way to get an upper bound. The idea is similar to #4 but use a set to store all possible power of 3 numbers (from 3^0 to 3^19). Say I want to know if "100000000000000000" is a power of 10. Initially, we will import the math module in the program. For example, if we call is_power_of_3(9), the function will return True because the logarithm of 9 with base 3 is 2, which is an integer. Everyone seems to have missed the fastest algorithm of all. JohnY, avoiding multiplication by dividing twice as often (and the fact that it doesn't handle zero) does not make for a good answer. Unexpected low characteristic impedance using the JLCPCB impedance calculator, Help Identify the name of the Hessen-Cassel Grenadier Company 1786. @HaoCS nice solution! Why are kiloohm resistors more used in op-amp circuits? You could be checking for integerness using x % 1 == 0 for integers, != 0 otherwise (thanks to @Peilonrayz in the comments): But, as noted in the comments, this works only until the precision of float is not enough anymore to distinguish that the result of the log is not an integer. Comments are not for extended discussion; this conversation has been, You can save a lot of work by using modulo exponentiation to check divisibility. rev2023.6.5.43475. +1 @Carl Smotzicz: The algorithm is inherently recursive, iteration just a workaround with no objective advantage (see tail-recursion elimination). Let's try to code to get the expected result with Python. has a certain beautiful simplicity to it (the last number is 3^19, so there aren't an absurd number of cases). Python program to check whether the number is a POWER of 3 or not. Typically your compiler will fold the constants for you, so there would be no loss of efficiency (that may not be the case in all languages/implementations). I really thought I could get this into the recursion limit, but it works for surprisingly big numbers. In your case, the worst case is that n = k^m, in which case it takes m steps. 1162261467 is the largest power of 3 that can fit into an Java int. This work is licensed under a Creative Commons Attribution 4.0 International License. There probably are a couple of taylor series hidden under the covers. from math import log def check_kth_power(n, k): return log(n, k) % 1 == 0 But, as noted in the comments, this works only until the precision of float is not enough anymore to distinguish that the result of the log is not an integer. With binary search, I'm sure this is the most efficient solution. Then it follows from the equation above that 3 | x if and only if 3 | (q + 2r). Second, the division introduces another rounding error. Suppose we have a number n. We have to check whether the number is the power of 3 or not. The deadline to purchase tickets is 9:45 p.m. Input Format: You will take three numbers as input from stdin, one on each line, respectively. Step 4: If this power when raised to three equals the number, n, return TRUE. at Facebook. OR, if you are using a system with, say, 64-bit integers, build a table of the powers of 3 and check against each entry in the array, it's only 40 elements. Which comes first: Continuous Integration/Continuous Delivery (CI/CD) or microservices? I find myself slightly thinking that if by 'integer' you mean 'signed 32-bit integer', then (pseudocode). Agree If the number can be reduced to 1, then the number is divisible by 3 else not. Really we will be doing a binary search on N, where 3^N = input value). However, if n is chosen at random, then the chances are (k-1) / k that you stop after just one step, (k-1) / k^2 for two steps, (k-1) / k^3 for three steps and so on. p log(3) = log(number) (taking log both side). The following tool visualize what the computer is doing step-by-step as it executes the said program: Contribute your code and comments through Disqus. The function first checks if the input is less than or equal to 0. 0 Answers Avg Quality 2/10 Grepper . 2. Bit of a wild guess, but perhaps they were looking for you to use divmod, since % and / with the same operands are essentially the same operation? Imho, this is too inefficent and imprecise, though mathematically correct. To get even more of a speed-boost, you can steal the nice premature exit from @pantarei's answer. Import the math module in the python program. Observation 1: If there is a power of three number, it will definitely end with either 3, 9 , 7 or 1. Sample Solution:- Python Code: import math def isPower (n, base): if base == 1 and n != 1: return False if base == 1 and n == 1: return True if base == 0 and n != 1: return False power = int (math.log(n, base) + 0.5) This is a summary of all good answers below this questions, and the performance figures can be found from the LeetCode article. speech to text on iOS continually makes same mistake, Help Identify the name of the Hessen-Cassel Grenadier Company 1786. Write a Python, C/C++ program to check if the given number is the power of 3 (k- any other integer number). You can do better than repeated division, which takes O(lg(X) * |division|) time. Now since we are clear with the observations, lets have a look at the algorithm. Note: Some number that is divisible by three not necessarily to be the power of three. Check if a given number N is a power of k Asked 5 years, 4 months ago Modified 3 years, 9 months ago Viewed 11k times 11 I am trying to solve a problem to check whether a given number N is a power of k. Example: Input 9, 3 > True // 3**2 = 9 Input 10, 2 > False Input 4096, 16 > True // 16**3 = 4096 I divide by 10^5 and get a quotient of 100 and a remainder of 0. And we have tocheck whether a number is a power of another number or not in Python by using max.log() function. In this program, we will use thepython while loop with function. What is the difficulty level of this exercise? And every power of an even is even. Python program to call a function before declaring it - Is it possible? Just using math.log, which is probably how you would do it in practice, also won't get your knowledge across. The Y = X case is elementary. These add to 10000000. 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 n= bx = bbb 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. In which jurisdictions is publishing false statements a codified crime? (Note 1) Keep in mind here the possibility that floating point imprecision may . Copyright Tuts Make . If I've put the notes correctly in the first piano roll image, why does it not sound correct? Let's start with a small example of finding the greatest number. Pardon my ignorance, but on the cycle counts, don't div and mul usually take more than one cycle? @Graipher Thanks, that's some interesting reading. My name is Devendra Dode. Here, the user will provide us two positive values a and b and we have to check whether a number is a power of another number or not in Python. To solve this problem simply, we will use the log() function from the math module.The math module provides us various mathematical operations and here we will use the log() function from this module. Two thirds of random input will be eliminated after a single mod and compare. This avoids division and keeps your firmly in the realm of integers. How to check if a number < 1 is power of 2? Shouldn't there be a check that the number is a positive value, though? This is one of the most common coding questions asked in a placement interview. You will be notified via email once the article is available for improvement. Affordable solution to train a team and make them project ready. Questions often give rise to new questions, but I recommend considering generalizations. Copyright 2023 www.includehelp.com. Write a Python program to check if a given positive integer is a power of three. I'm pretty sure they wanted you to use binary search. This solution is specific to C++11, using template meta programming so that complier will replace the call isPowerOf3
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