Skip to content
Related Articles

Related Articles

Improve Article

Find whether a given integer is a power of 3 or not

  • Difficulty Level : Easy
  • Last Updated : 04 Oct, 2021
Geek Week

Given a positive integer, write a function to find if it is a power of three or not. 
Examples: 

Input : 3
Output :Yes

Input :6
Output :No

Recursive approach :

Check if the number is divisible by 3, if yes then keep checking the same for number/3 recursively. If the number can be reduced to 1, then the number is divisible by 3 else not.

C++




#include <bits/stdc++.h>
#define ll long long
using namespace std;
bool isPower_of_Three(ll n)
{
    if (n <= 0)
        return false;
    if (n % 3 == 0)
        return isPower_of_Three(n / 3);
    if (n == 1)
        return true;
    return false;
}
int main()
{
    ll num1;
    num1 = 243;
    if (isPower_of_Three(num1))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    ll num2 = 6;
    if (isPower_of_Three(num2))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

Java




import java.util.*;
 
class GFG{
static boolean isPower_of_Three(long n)
{
    if (n <= 0)
        return false;
    if (n % 3 == 0)
        return isPower_of_Three(n / 3);
    if (n == 1)
        return true;
    return false;
}
   
  // Driver code
public static void main(String[] args)
{
    long  num1 = 243;
    if (isPower_of_Three(num1))
        System.out.print("Yes" +"\n");
    else
        System.out.print("No" +"\n");
    long num2 = 6;
    if (isPower_of_Three(num2))
        System.out.print("Yes" +"\n");
    else
        System.out.print("No" +"\n");
}
}
 
// This code is contributed by umadevi9616

Python3




def isPower_of_Three(n):
 
    if (n <= 0):
        return False
    if (n % 3 == 0):
        return isPower_of_Three(n // 3)
    if (n == 1):
        return True
    return False
 
 # Driver code
num1 = 243
if (isPower_of_Three(num1)):
    print("Yes")
else:
    print("No")
     
num2 = 6
if (isPower_of_Three(num2)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by shivanisinghss2110

C#




using System;
 
class GFG{
static Boolean isPower_of_Three(long n)
{
    if (n <= 0)
        return false;
    if (n % 3 == 0)
        return isPower_of_Three(n / 3);
    if (n == 1)
        return true;
    return false;
}
   
  // Driver code
public static void Main(String[] args)
{
    long  num1 = 243;
    if (isPower_of_Three(num1))
        Console.Write("Yes" +"\n");
    else
        Console.Write("No" +"\n");
    long num2 = 6;
    if (isPower_of_Three(num2))
        Console.Write("Yes" +"\n");
    else
        Console.Write("No" +"\n");
}
}
 
// this code is contributed by shivanisinghss2110

Javascript




<script>
function isPower_of_Three(n)
{
    if (n <= 0)
        return false;
    if (n % 3 == 0)
        return isPower_of_Three(n / 3);
    if (n == 1)
        return true;
    return false;
}
 
     
    let num1 = 243;
    if (isPower_of_Three(num1))
      document.write("Yes");
    else
          document.write("No");
    let num2 = 6;
    if (isPower_of_Three(num2))
      document.write("Yes");
    else
        document.write("</br>No");
         
        //This code is contributed by vaibhavrabadiyaa3.
</script>
Output
Yes
No

Approach:
The logic is very simple. 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. 



C++




// C++ program to check if a number is power
// of 3 or not.
#include <iostream>
using namespace std;
 
// Returns true if n is power of 3, else false
bool check(int n)
{
    if (n <= 0)
        return false;
   
    /* The maximum power of 3 value that
       integer can hold is 1162261467 ( 3^19 ) .*/
    return 1162261467 % n == 0;
}
 
// Driver code
int main()
{
    int n = 9;
    if (check(n))
        cout <<"Yes";
    else
        cout <<"No";
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C




// C++ program to check if a number is power
// of 3 or not.
#include <stdio.h>
#include <stdbool.h>
 
// Returns true if n is power of 3, else false
bool check(int n)
{
    if (n <= 0)
        return false;
    /* The maximum power of 3 value that
       integer can hold is 1162261467 ( 3^19 ) .*/
    return 1162261467 % n == 0;
}
 
// Driver code
int main()
{
    int n = 9;
    if (check(n))
        printf("Yes");
    else
        printf("No");
 
    return 0;
}

Java




// Java program to check if a number is power
// of 3 or not.
public class Power_3 {
 
    // Returns true if n is power of 3, else false
    static boolean check(int n)
    {
        /* To prevent
        java.lang.ArithmeticException: / by zero and
        negative n */
        if (n <= 0)
            return false;
        /* The maximum power of 3 value that
           integer can hold is 1162261467 ( 3^19 ) .*/
        return 1162261467 % n == 0;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 9;
        if (check(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Sumit Ghosh

Python




# Python program to check if a number is power
# of 3 or not.
  
# Returns true if n is power of 3, else false
def check(n):
    """ The maximum power of 3 value that
       integer can hold is 1162261467 ( 3^19 ) ."""
    return 1162261467 % n == 0
  
# Driver code
n = 9
if (check(n)):
    print ("Yes")
else:
    print ("No")
 
# This code is contributed by Sachin Bisht

C#




// C# program to check if a number
// is power of 3 or not.
using System;
 
public class GFG {
 
    // Returns true if n is power
    // of 3, else false
    static bool check(int n)
    {
        if (n <= 0)
            return false;
        /* The maximum power of 3
        value that integer can hold
        is 1162261467 ( 3^19 ) .*/
        return 1162261467 % n == 0;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 9;
        if (check(n))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed by
// nitin mittal.

PHP




<?php
// PHP program to check if a
// number is power of 3 or not.
 
// Returns true if n is
// power of 3, else false
function check($n)
{
     
    /* The maximum power of 3 value that
       integer can hold is 1162261467
       ( 3^19 ) . */
    return 1162261467 % $n == 0;
}
 
    // Driver code
    $n = 9;
    if (check($n))
        echo("Yes");
    else
        echo("No");
 
// This code is contributed by nitin mittal
?>

Javascript




<script>
// Javascript program to check if a
// number is power of 3 or not.
 
// Returns true if n is
// power of 3, else false
function check(n)
{
     
    /* The maximum power of 3 value that
    integer can hold is 1162261467
    ( 3^19 ) . */
    return 1162261467 % n == 0;
}
 
    // Driver code
    let n = 9;
    if (check(n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by nitin _saurabh_jaiswal
</script>
Output
Yes

Time Complexity : O(1)

Auxiliary Space: O(1)
This article is contributed by Jebasingh and Riyazath

Approach:

This approach is based on the below simple observations.

Observation 1: If there is a power of three number, it will definitely end with either 3, 9 , 7 or 1.

Observation 2 : If a number ends with one of these 4 digits, we only have to check the powers of three which would guarantee a number ending with that last digit. 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.

Now since we are clear with the observations, let’s have a look at the algorithm.



Algorithm :  

Step 1: If the given number, n, is not ending with 3,9,7 or 1, it means that the number is not a power of three, therefore return FALSE.

Step 2 : If not, we create a Map with 4 entries in it in order to maintain the mapping between the powers to three(1,2,3,4) and the number’s last digits(3,9,7,1).

Step 3 : Extract the last digit from a given number and look up it’s corresponding power in the map.

Step 4 :  If this power when raised to three equals  the number, n, return TRUE.

Step 5 : If this power raised to three is less than the number, n, increment the power straight by 4 and loop step 4 until the power raised to three becomes more than n.  

Step 6 : If the power raised to three becomes more than the given number, return FALSE.

C++




#include <bits/stdc++.h>
using namespace std;
bool isPowerOfThree(int n)
{
    if (n == 1)
        return true;
    int lastDigit = n % 10;
    map<int, int> map;
    map[3] = 1;
    map[9] = 2;
    map[7] = 3;
    map[1] = 4;
 
    if (!map[lastDigit])
        return false;
 
    int power = map[lastDigit];
    double powerOfThree = pow(3, power);
    while (powerOfThree <= n) {
        if (powerOfThree == n)
            return true;
        power = power + 4;
        powerOfThree = pow(3, power);
    }
    return false;
}
int main()
{
    int n = 81;
    cout << (isPowerOfThree(n) ? "true" : "false") << endl;
    n = 91;
    cout << (isPowerOfThree(n) ? "true" : "false") << endl;
    return 0;
}
 
// This code is contributed by umadevi9616

Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.*;
 
class GFG {
    public static boolean isPowerOfThree(int n)
    {
        if (n == 1)
            return true;
        int lastDigit = n % 10;
        Map<Integer, Integer> map = new HashMap<>();
        map.put(3, 1);
        map.put(9, 2);
        map.put(7, 3);
        map.put(1, 4);
 
        if (map.get(lastDigit) == null)
            return false;
 
        int power = map.get(lastDigit);
        double powerOfThree = Math.pow(3, power);
        while (powerOfThree <= n) {
            if (powerOfThree == n)
                return true;
            power = power + 4;
            powerOfThree = Math.pow(3, power);
        }
        return false;
    }
    public static void main(String[] args)
    {
        int n = 81;
        System.out.println(isPowerOfThree(n));
        n = 91;
        System.out.println(isPowerOfThree(n));
    }
}

C#




/*package whatever //do not write package name here */
using System;
using System.Collections.Generic;
 
public class GFG {
    public static bool isPowerOfThree(int n)
    {
        if (n == 1)
            return true;
        int lastDigit = n % 10;
        Dictionary<int, int> map = new Dictionary<int,int>();
        map.Add(3, 1);
        map.Add(9, 2);
        map.Add(7, 3);
        map.Add(1, 4);
 
        if (!map.ContainsValue(lastDigit))
            return false;
 
        int power = map[lastDigit];
        double powerOfThree = Math.Pow(3, power);
        while (powerOfThree <= n) {
            if (powerOfThree == n)
                return true;
            power = power + 4;
            powerOfThree = Math.Pow(3, power);
        }
        return false;
    }
    public static void Main(String[] args)
    {
        int n = 81;
        Console.WriteLine(isPowerOfThree(n));
        n = 91;
        Console.WriteLine(isPowerOfThree(n));
    }
}
 
// This code is contributed by umadevi9616

Javascript




<script>
/*package whatever //do not write package name here */   
function isPowerOfThree(n) {
        if (n == 1)
            return true;
        var lastDigit = n % 10;
        var map = new Map();
        map.set(3, 1);
        map.set(9, 2);
        map.set(7, 3);
        map.set(1, 4);
 
        if (map.get(lastDigit) == null)
            return false;
 
        var power = map.get(lastDigit);
        var powerOfThree = Math.pow(3, power);
        while (powerOfThree <= n) {
            if (powerOfThree == n)
                return true;
            power = power + 4;
            powerOfThree = Math.pow(3, power);
        }
        return false;
    }
 
    // Driver code
        var n = 81;
        document.write(isPowerOfThree(n)+"<br/>");
        n = 91;
        document.write(isPowerOfThree(n));
 
// This code is contributed by umadevi9616
</script>
Output
true
false

Analysis:

Runtime Complexity:

O(1) : Since the given number is an Integer, it can at max be 2147483647 (32 bit) and the highest power of three that is less than or equal to this number is 3^19 = 1162261467. And since we increment the power by 4, we will have a loop running at most 5 times, hence O(1).

Space Complexity:

O(1) : Since we only have 4 entries in a Map no matter how big the number is given to us.

This approach is contributed by Durgesh Valecha.

If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :