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# Logarithm tricks for Competitive Programming

• Difficulty Level : Medium
• Last Updated : 01 Sep, 2022

Logarithm: is the inverse function of the exponentiation which means the logarithm value of a given number x is the exponent to another number.

Below are some tricks using Logarithmic function which can be handy in competitive programming.

### Checking if a number is a power of two or not:

Given an integer N, the task is to check that if the number N is the power of 2.

Examples:

Input: N = 8
Output: Yes

Input: N = 6
Output: No

Approach: A simple method for this is to simply take the log of the number on base 2, if you get an integer then the number is the power of 2.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check that // a integer is a power of Two   #include    using namespace std;   // Function to check if the number // is a power of two bool isPowerOfTwo(int n) {     return (ceil(log2(n)) == floor(log2(n))); }   // Driver Code int main() {     int N = 8;       if (isPowerOfTwo(N)) {         cout << "Yes";     }     else {         cout << "No";     } }

## C

 // C implementation to check that  // a integer is a power of Two  #include  #include    // Function to check if the number  // is a power of two  _Bool isPowerOfTwo(int n)  {      return (ceil(log2(n)) == floor(log2(n)));  }    // Driver Code  int main()  {      int N = 8;        if (isPowerOfTwo(N))     {          printf("Yes");      }      else     {          printf("No");      }  }    // This code is contributed by vikas_g

## Java

 // Java implementation to check that  // a integer is a power of Two  import java.lang.Math;   class GFG{       // Function to check if the number  // is a power of two  public static boolean isPowerOfTwo(int n)  {      return(Math.ceil(Math.log(n) /                      Math.log(2)) ==            Math.floor(Math.log(n) /                       Math.log(2))); }        // Driver Code  public static void main(String[] args) {     int N = 8;        if (isPowerOfTwo(N))      {          System.out.println("Yes");     }      else     {         System.out.println("No");     }  } }   // This code is contributed by divyeshrabadiya07

## Python3

 # Python3 implementation to check that  # a integer is a power of two  import math   # Function to check if the number  # is a power of two              def isPowerOfTwo(n):           return(math.ceil(math.log(n) //                      math.log(2)) ==            math.floor(math.log(n) //                       math.log(2)));                        # Driver code  if __name__=='__main__':           N = 8           if isPowerOfTwo(N):         print('Yes')     else:         print('No')   # This code is contributed by rutvik_56

## C#

 // C# implementation to check that  // a integer is a power of Two  using System;   class GFG{       // Function to check if the number  // is a power of two  public static bool isPowerOfTwo(int n)  {      return(Math.Ceiling(Math.Log(n) /                          Math.Log(2)) ==               Math.Floor(Math.Log(n) /                         Math.Log(2))); }        // Driver Code  public static void Main(String[] args) {     int N = 8;        if (isPowerOfTwo(N))      {          Console.WriteLine("Yes");     }      else     {         Console.WriteLine("No");     }  } }   // This code is contributed by 29AjayKumar

## Javascript

 

Output:

Yes

Time Complexity: O(logn)

Auxiliary Space: O(1)

### Kth root of a Number

Given two integers N and K, the task is to find the Kth root of the number N.

Examples:

Input: N = 8, K = 3
Output: 2

Input: N = 32, K = 5
Output:

Approach: A simple solution is to use logarithmic function to find the Kth root of the number. Below is the illustration of the approach:

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Applying on both side
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=>
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Below is the implementation of the above approach:

## C++

 // C++ implementation to find // Kth root of the number   #include    using namespace std;   // Function to find the // Kth root of the number double kthRoot(double n, int k) {     return pow(k,                (1.0 / k)                    * (log(n)                       / log(k))); }   // Driver Code int main() {     double N = 8.0;     int K = 3;       cout << kthRoot(N, K);       return 0; }

## Java

 // Java implementation to find // Kth root of the number class GFG{   // Function to find the // Kth root of the number static double kthRoot(double n, int k) {     return Math.pow(k, (1.0 / k) *                      (Math.log(n) / Math.log(k))); }   // Driver Code public static void main(String[] args) {     double N = 8.0;     int K = 3;       System.out.print(kthRoot(N, K)); } }   // This code is contributed by shivanisinghss2110

## Python3

 # Python3 implementation to find  # Kth root of the number  import math   # Function to find the  # Kth root of the number  def kth_root(n, k):           return(pow(k, ((1.0 / k) * (math.log(n) /                                 math.log(k)))))   # Driver code if __name__=="__main__":           n = 8.0     k = 3           print(round(kth_root(n, k)))   # This code is contributed by dipesh99kumar

## C#

 // C# implementation to find  // Kth root of the number using System;   // Function to find the  // Kth root of the number class GFG{       static double kthRoot(double n, int k)  {      return Math.Pow(k, (1.0 / k) *                     (Math.Log(n) / Math.Log(k)));  }   // Driver Code  public static void Main() {     double N = 8.0;      int K = 3;        Console.Write(kthRoot(N, K));  } }   // This code is contributed by vikas_g

## Javascript

 

Output:

2

Time Complexity: O(logn)

Auxiliary Space: O(1)

### Count digits in a Number:

Given an integer N, the task is to count the digits in a number N.

Examples:

Input: N = 243
Output: 3

Input: N = 1000
Output:

Approach: The idea is to find the logarithm of the number base 10 to count the number of digits.

Below is the implementation of the above approach:

## C++

 // C++ implementation count the // number of digits in a number   #include    using namespace std;   // Function to count the // number of digits in a number int countDigit(long long n) {     return floor(log10(n) + 1); }   // Driver Code int main() {     double N = 80;       cout << countDigit(N);       return 0; }

## C

 // C implementation count the  // number of digits in a number  #include  #include    // Function to count the  // number of digits in a number  int countDigit(long long n)  {      return (floor(log10(n) + 1));  }    // Driver Code  int main()  {      double N = 80;        printf("%d", countDigit(N));        return 0;  }    // This code is contributed by vikas_g

## Java

 // Java implementation to count the  // number of digits in a number class GFG{    // Function to count the  // number of digits in a number  static int countDigit(double n)  {      return((int)Math.floor(Math.log10(n) + 1));   }    // Driver Code  public static void main(String[] args)  {      double N = 80;            System.out.println(countDigit(N));  }  }    // This code is contributed by vikas_g

## Python3

 # Python3 implementation count the  # number of digits in a number import math   # Function to count the  # number of digits in a number  def countDigit(n):           return(math.floor(math.log10(n) + 1))   # Driver code if __name__=="__main__":           n = 80       print(countDigit(n))   # This code is contributed by dipesh99kumar

## C#

 // C# implementation count the  // number of digits in a number using System;   // Function to count the  // number of digits in a number class GFG{       static int countDigit(double n)  {      return((int)Math.Floor(Math.Log10(n) + 1)); }   // Driver Code  public static void Main() {     double N = 80;            Console.Write(countDigit(N));  } }   // This code is contributed by vikas_g

## Javascript

 

Output:

2

Time Complexity: O(logn)

Auxiliary Space: O(1)

### Check if N is a power of K or not:

Given two integers N and K, the task is to check if Y is power of X or not.

Examples:

Input: N = 8, K = 2
Output: Yes

Input: N = 27, K = 3
Output: Yes

Approach: The idea is to take log of N in base K. If it turns out to be an integer, then N is a power of K.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check if // the number is power of K   #include    using namespace std;   // Function to check if // the number is power of K bool isPower(int N, int K) {     // logarithm function to     // calculate value     int res1 = log(N) / log(K);     double res2 = log(N) / log(K);       // compare to the result1     // or result2 both are equal     return (res1 == res2); }   // Driver Code int main() {     int N = 8;     int K = 2;       if (isPower(N, K)) {         cout << "Yes";     }     else {         cout << "No";     }       return 0; }

## C

 // C implementation to check if  // the number is power of K  #include  #include    // Function to check if  // the number is power of K  _Bool isPower(int N, int K)  {            // Logarithm function to      // calculate value      int res1 = log(N) / log(K);      double res2 = log(N) / log(K);        // Compare to the result1      // or result2 both are equal      return (res1 == res2);  }    // Driver Code  int main()  {      int N = 8;      int K = 2;        if (isPower(N, K))     {          printf("Yes");      }      else     {          printf("No");      }      return 0;  }    // This code is contributed by vikas_g

## Java

 // Java implementation to check if  // the number is power of K  class GFG{    // Function to check if  // the number is power of K  static boolean isPower(int N, int K)  {            // Logarithm function to      // calculate value      int res1 = (int)(Math.log(N) / Math.log(K));      double res2 = Math.log(N) / Math.log(K);        // Compare to the result1      // or result2 both are equal      return (res1 == res2);  }    // Driver Code  public static void main(String[] args)  {      int N = 8;      int K = 2;        if (isPower(N, K))      {          System.out.println("Yes");      }      else     {          System.out.println("No");      } }  }    // This code is contributed by vikas_g

## Python3

 # Python3 implementation to check if a  # number is a power of the other number from math import log   # Function to check if  # the number is power of K  def isPower(n, k):           # Logarithm function to      # calculate value      res1 = int(log(n) / log(k))     res2 = log(n) / log(k)           # Compare to the result1      # or result2 both are equal      return(res1 == res2)   # Driver code if __name__=="__main__":           n = 8     k = 2           if (isPower(n, k)):         print("Yes")     else:         print("No")   # This code is contributed by dipesh99kumar

## C#

 // C# implementation to check if // the number is power of K using System;   // Function to count the  // number of digits in a number class GFG{       static bool isPower(int N, int K)  {            // Logarithm function to      // calculate value      int res1 = (int)(Math.Log(N) / Math.Log(K));      double res2 = Math.Log(N) / Math.Log(K);        // Compare to the result1      // or result2 both are equal      return (res1 == res2); }   // Driver Code  public static void Main() {     int N = 8;      int K = 2;        if (isPower(N, K))      {          Console.Write("Yes");      }      else     {          Console.Write("No");      } } }   // This code is contributed by vikas_g

## Javascript

 

Output:

Yes

Time Complexity: O(logn)

Auxiliary Space: O(1)

### To find the power of K greater than equal to and less than equal to N:

Given two integers N and K, the task is to find the power of K greater than equal to and less than equal to N.

Examples:

Input: N = 7, K = 2
Output: 4 8

Input: N = 18, K = 3
Output: 9 27

Approach: The idea is to find the floor value of the log K value of the given integer N, Then compute the Kth power of this number to compute the previous and next Kth power.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the // previous and next power of K   #include    using namespace std;   // Function to return the highest power // of k less than or equal to n int prevPowerofK(int n, int k) {     int p = (int)(log(n) / log(k));     return (int)pow(k, p); }   // Function to return the smallest power // of k greater than or equal to n int nextPowerOfK(int n, int k) {     return prevPowerofK(n, k) * k; }   // Driver Code int main() {     int N = 7;     int K = 2;       cout << prevPowerofK(N, K) << " ";       cout << nextPowerOfK(N, K) << endl;     return 0; }

## C

 // C implementation to find the  // previous and next power of K #include  #include    // Function to return the highest power  // of k less than or equal to n  int prevPowerofK(int n, int k)  {      int p = (int)(log(n) / log(k));      return (int)pow(k, p);  }    // Function to return the smallest power  // of k greater than or equal to n  int nextPowerOfK(int n, int k)  {      return prevPowerofK(n, k) * k;  }    // Driver Code  int main()  {      int N = 7;      int K = 2;        printf("%d ", prevPowerofK(N, K));      printf("%d\n", nextPowerOfK(N, K));            return 0;  }    // This code is contributed by vikas_g

## Java

 // Java implementation to find the  // previous and next power of K  class GFG{    // Function to return the highest power  // of k less than or equal to n  static int prevPowerofK(int n, int k)  {      int p = (int)(Math.log(n) / Math.log(k));      return (int)Math.pow(k, p);  }    // Function to return the smallest power  // of k greater than or equal to n  static int nextPowerOfK(int n, int k)  {      return prevPowerofK(n, k) * k;  }    // Driver Code  public static void main(String[] args)  {      int N = 7;      int K = 2;            System.out.print(prevPowerofK(N, K) + " ");      System.out.println(nextPowerOfK(N, K));  }  }    // This code is contributed by vikas_g

## Python3

 # Python3 implementation to find the # previous and next power of K  from math import log   # Function to return the highest power # of k less than or equal to n def prevPowerofK(n, k):           p = (int)(log(n) / log(k));     return pow(k, p);   # Function to return the smallest power # of k greater than or equal to n def nextPowerOfK(n, k):           return prevPowerofK(n, k) * k;   # Driver Code if __name__=="__main__":           N = 7     K = 2           print(prevPowerofK(N, K), end = " ")     print(nextPowerOfK(N, K))   # This code is contributed by dipesh99kumar

## C#

 // C# implementation to find the  // previous and next power of K using System;   // Function to count the  // number of digits in a number class GFG{       // Function to return the highest power  // of k less than or equal to n  static int prevPowerofK(int n, int k)  {      int p = (int)(Math.Log(n) / Math.Log(k));      return (int)Math.Pow(k, p);  }    // Function to return the smallest power  // of k greater than or equal to n  static int nextPowerOfK(int n, int k)  {      return prevPowerofK(n, k) * k;    }   // Driver Code  public static void Main() {     int N = 7;      int K = 2;            Console.Write(prevPowerofK(N, K) + " ");      Console.Write(nextPowerOfK(N, K));  } }   // This code is contributed by vikas_g

## Javascript

 

Output:

4 8

Time Complexity: O(logn)

Auxiliary Space: O(1)

### To Find the position of rightmost set bit:

Given an integer N, the task is to find the position of the rightmost set bit.

Examples:

Input: N = 7
Output: 1

Input: N = 8
Output:

Approach:

• Take two’s complement of the given no as all bits are reverted except the first ‘1’ from right to left (0111)
• Do a bit-wise & with original no, this will return no with the required one only (0100)
• Take the log2 of the no, you will get (position – 1) (2)

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the // rightmost set bit   #include    using namespace std;   // Function to find the rightmost // bit set of the integer N unsigned int getFirstSetBitPos(int n) {     return log2(n & -n) + 1; }   // Driver Code int main() {     int N = 8;       cout << getFirstSetBitPos(N);     return 0; }

## C

 // C implementation to find the  // rightmost set bit  #include  #include    // Function to find the rightmost  // bit set of the integer N  unsigned int getFirstSetBitPos(int n)  {      return log2(n & -n) + 1;  }    // Driver Code  int main()  {      int N = 8;        printf("%d", getFirstSetBitPos(N));      return 0;  }    // This code is contributed by vikas_g

## Java

 // Java implementation to find the  // rightmost set bit  class GFG{    // Function to find the rightmost  // bit set of the integer N  static int getFirstSetBitPos(int n)  {      return (int)(Math.log(n & -n) /                   Math.log(2)) + 1;  }    // Driver Code  public static void main(String[] args)  {      int N = 8;        System.out.println(getFirstSetBitPos(N));  }  }    // This code is contributed by vikas_g

## Python3

 # Python3 implementation to find the # rightmost set bit import math   # Function to find the rightmost # bit set of the integer N def getFirstSetBitPos(n):       return math.log2(n & -n) + 1;   # Driver Code if __name__=="__main__":           N = 8       print(int(getFirstSetBitPos(N)))   # This code is contributed by dipesh99kumar

## C#

 // C# implementation to find the  // rightmost set bit  using System;   class GFG{   // Function to find the rightmost  // bit set of the integer N  static int getFirstSetBitPos(int n)  {      return (int)(Math.Log(n & -n) /                   Math.Log(2)) + 1; }   // Driver Code  public static void Main() {     int N = 8;        Console.Write(getFirstSetBitPos(N));  } }   // This code is contributed by vikas_g

## Javascript

 

Output:

4

Time Complexity: O(logn)

Auxiliary Space: O(1)

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