Count number of 0s in base K representation of a number

Given a number N, the task is to find the number of zero's in base K representation of the given number, where K > 1.

Examples:

Input: N = 10, K = 3
Output: 1
Explanation: Base 3 representation of 10 is 101. 
Hence the number of 0's in 101 is 1.

Input: N = 8, K = 2
Output: 3
Explanation: Base 2 representation of 8 is 1000. 
Hence the number of 0's in 1000 is 3.

Approach: This problem can be solved by converting the number into base K form. 
The pseudocode for converting a number N of base 10 to base K :

allDigits = ""
While N > 0:
       digit = N%k
       allDigits.append(digit)
       N /= k
number_in_base_K = reversed(allDigits)

Follow the steps mentioned below to implement the approach:

• Start forming the K base number as shown in the above pseudocode.
• Instead of forming the number only check in each pass of the loop is the current digit being formed is 0 or not to.
• Return the total count of 0s.

Below is the implementation of the above approach.

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to count the number of zeros
// in K base representation of N
int countZero(int N, int K)
{
    // Making a variable to store count
    int count = 0;

    // Looping till n becomes 0
    while (N > 0) {

        // Check if the current digit
        // is 0 or not.
        if (N % K == 0)
            count++;
        N /= K;
    }
    return count;
}

// Driver code
int main()
{
    int N = 8;
    int K = 2;

    // Function call
    cout << countZero(N, K) << endl;
    return 0;
}

// This code is contributed by Ashish Kumar

// C code to implement the approach
#include<stdio.h>

// Function to count the number of zeros
// in K base representation of N
int countZero(int N, int K)
{
  
  // Making a variable to store count
  int count = 0;

  // Looping till n becomes 0
  while (N > 0) {

    // Check if the current digit
    // is 0 or not.
    if (N % K == 0)
      count++;
    N /= K;
  }
  return count;
}

// Driver code
int main()
{
  int N = 8;
  int K = 2;

  // Function call
  int ans = countZero(N,K);
  printf("%d",ans);
  return 0;
}

// This code is contributed by ashishsingh13122000.

// Java code to implement the approach
import java.io.*;

class GFG 
{

  // Function to count the number of zeros
  // in K base representation of N
  public static int countZero(int N, int K)
  {

    // Making a variable to store count
    int count = 0;

    // Looping till n becomes 0
    while (N > 0) {

      // Check if the current digit
      // is 0 or not.
      if (N % K == 0)
        count++;
      N /= K;
    }
    return count;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 8;
    int K = 2;

    // Function call
    System.out.println(countZero(N, K));
  }
}

// This code is contributed by Rohit Pradhan

# Python code to implement the approach

# Function to count the number of zeros
# in K base representation of N
def countZero(N, K):
  
    # Making a variable to store count
    count = 0

    # Looping till n becomes 0
    while (N > 0):

        # Check if the current digit
        # is 0 or not.
        if (N % K == 0):
            count += 1
        N //= K

    return count

# Driver code
N = 8
K = 2

# Function call
print(countZero(N, K))

# This code is contributed by shinjanpatra

// C# code to implement the approach
using System;

class GFG {

  // Function to count the number of zeros
  // in K base representation of N
  static int countZero(int N, int K)
  {

    // Making a variable to store count
    int count = 0;

    // Looping till n becomes 0
    while (N > 0) {

      // Check if the current digit
      // is 0 or not.
      if (N % K == 0)
        count++;
      N /= K;
    }
    return count;
  }

  // Driver Code
  public static void Main()
  {
    int N = 8;
    int K = 2;

    // Function call
    Console.Write(countZero(N, K));
  }
}

// This code is contributed by Samim Hossain Mondal.

  <script>
        // JavaScript code for the above approach

        // Function to count the number of zeros
        // in K base representation of N
        function countZero(N, K)
        {
        
            // Making a variable to store count
            let count = 0;

            // Looping till n becomes 0
            while (N > 0) {

                // Check if the current digit
                // is 0 or not.
                if (N % K == 0)
                    count++;
                N = Math.floor(N / K);
            }
            return count;
        }

        // Driver code
        let N = 8;
        let K = 2;

        // Function call
        document.write(countZero(N, K) + '<br>');

    // This code is contributed by Potta Lokesh
    </script>

3

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

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Article Tags :

• Bit Magic
• Mathematical
• DSA

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• Bit Magic
• Mathematical