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Count bases which contains a set bit as the Most Significant Bit in the representation of N

  • Last Updated : 08 Jul, 2021

Given a positive integer N, the task is to count the number of different bases in which, when N is represented, the most significant bit of N is a found to be a set bit.

Examples:

Input: N = 6
Output: 4
Explanation: The number 6 can be represented in 5 different bases, i.e. base 2, base 3, base 4, base 5, base 6.

  1. (6)10  in base 2: (110)2
  2. (6)10  in base 3: (20)3
  3. (6)10  in base 4: (12)4
  4. (6)10  in base 5: (11)5
  5. (6)10  in base 6: (10)6

The base representation for (6)10 in base 2, base 4, base 5, base 6 starts with 1. Hence, the required count of bases is 4.

Input: N = 5
Output: 4



Approach: The given problem can be solved by finding the MSB of the given number N for every possible base and count those bases that have MSB as 1. Follow the steps below to solve the problem:

  • Initialize a variable, say count as 0, to store the required result.
  • Iterate over the range [2, N] using a variable, say B, and perform the following steps:
    • Store the highest power of base B required to represent number N in a variable P. This can be easily achieved by finding the value of (log N to the base B) i.e., logB(N) truncated to the nearest integer.
    • To find the value of logB(N) use the log property: logB(N) = log(N)/log(B)
    • Store the most significant digit of N by dividing N by (B)P. If it is equal to 1, then increment the value of the count by 1.
  • After completing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count bases
// having MSB of N as a set bit
int countOfBase(int N)
{
    // Store the required count
    int count = 0;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; ++i) {
 
        int highestPower
            = (int)(log(N) / log(i));
 
        // Store the MSB of N
        int firstDigit = N / (int)pow(
                                 i, highestPower);
 
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1) {
            ++count;
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 6;
    cout << countOfBase(N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
class GFG {
     
 public static int countOfBase(int N)
{
    
    // Store the required count
    int count = 0;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; ++i) {
 
        int highestPower
            = (int)(Math.log(N) /Math.log(i));
 
        // Store the MSB of N
        int firstDigit = N / (int)Math.pow(
                                 i, highestPower);
 
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1) {
            ++count;
        }
    }
 
    // Return the count
    return count;
}
 
  // DRIVER CODE
    public static void main (String[] args)
    {
       int N = 6;
 
        System.out.println(countOfBase(N));
    }
}
 
 // This code is contributed by Potta Lokesh

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count bases
// having MSB of N as a set bit
static int countOfBase(int N)
{
     
    // Store the required count
    int count = 0;
     
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; ++i)
    {
     
        int highestPower = (int)(Math.Log(N) /
                                 Math.Log(i));
         
        // Store the MSB of N
        int firstDigit = N / (int)Math.Pow(
                                 i, highestPower);
         
        // If MSB is 1, then increment
        // the count by 1
        if (firstDigit == 1)
        {
            ++count;
        }
    }
     
    // Return the count
    return count;
}
 
// Driver Code
public static void Main()
{
    int N = 6;
     
    Console.Write(countOfBase(N));
}
}
 
// This code is contributed by ipg2016107

Javascript




<script>
        // JavaScript implementation of the approach
        function countOfBase(N)
        {
         
            // Store the required count
            let count = 0;
 
            // Iterate over the range [2, N]
            for (let i = 2; i <= N; ++i) {
 
                let highestPower
                    = parseInt(Math.log(N) / Math.log(i));
 
                // Store the MSB of N
                let firstDigit = parseInt(N / Math.pow(
                    i, highestPower));
 
                // If MSB is 1, then increment
                // the count by 1
                if (firstDigit == 1) {
                    ++count;
                }
            }
 
            // Return the count
            return count;
        }
         
        // Driver code
        let N = 6;
        document.write(countOfBase(N))
 
// This code is contributed by Potta Lokesh
 
 
    </script>

Python3




# Python 3 program for the above approach
import math
 
# Function to count bases
# having MSB of N as a set bit
def countOfBase(N) :
     
    # Store the required count
    count = 0
 
    # Iterate over the range [2, N]
    for i in range(2, N+1):
 
        highestPower = int(math.log(N) / math.log(i))
 
        # Store the MSB of N
        firstDigit = int(N / int(math.pow(i, highestPower)))
 
        # If MSB is 1, then increment
        # the count by 1
        if (firstDigit == 1) :
            count += 1
         
     
 
    # Return the count
    return count
 
 
# Driver Code
 
N = 6
print(countOfBase(N))
Output: 
4

 

Time Complexity: O(N)
Auxiliary Space: O(1)

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