Find nth Magic Number

2.7

A magic number is defined as a number which can be expressed as a power of 5 or sum of unique powers of 5. First few magic numbers are 5, 25, 30(5 + 25), 125, 130(125 + 5), ….

Write a function to find the nth Magic number.

Example:

Input: n = 2
Output: 25

Input: n = 5
Output: 130 

Source: Amazon written round question

If we notice carefully the magic numbers can be represented as 001, 010, 011, 100, 101, 110 etc, where 001 is 0*pow(5,3) + 0*pow(5,2) + 1*pow(5,1). So basically we need to add powers of 5 for each bit set in given integer n.

Below is the implementation based on this idea.

CPP

// C++ program to find nth magic numebr
#include <bits/stdc++.h>
using namespace std;

// Function to find nth magic numebr
int nthMagicNo(int n)
{
    int pow = 1, answer = 0;

    // Go through every bit of n
    while (n)
    {
       pow = pow*5;

       // If last bit of n is set
       if (n & 1)
         answer += pow;

       // proceed to next bit
       n >>= 1;  // or n = n/2
    }
    return answer;
}

// Driver program to test above function
int main()
{
    int n = 5;
    cout << "nth magic number is " << nthMagicNo(n) << endl;
    return 0;
}

Java

// Java program to find nth
// magic numebr
import java.io.*;

class GFG 
{
  // Function to find nth magic number
  static int nthMagicNo(int n)
  {
     int pow = 1, answer = 0;
 
     // Go through every bit of n
     while (n != 0)
     {
       pow = pow*5;
 
       // If last bit of n is set
       if ((int)(n & 1) == 1)
         answer += pow;
 
       // proceed to next bit
       // or n = n/2
       n >>= 1;  
     }
     return answer;
  }
 
  // Driver program to test
  // above function
  public static void main(String[] args)
  {
    int n = 5;
    System.out.println("nth magic" +
    " number is " + nthMagicNo(n));
  }
}


// This code is contributed by
// prerna saini

Python3

# Python program to find nth magic numebr

# Function to find nth magic numebr
def nthMagicNo(n):

    pow = 1
    answer = 0

    # Go through every bit of n
    while (n):

        pow = pow*5

        # If last bit of n is set
        if (n & 1):
            answer += pow

        # proceed to next bit
        n >>= 1 # or n = n/2
    
    return answer


# Driver program to test above function
n = 5
print("nth magic number is", nthMagicNo(n))

# This code is contributed by
# Smitha Dinesh Semwal


Output:
 nth magic number is 130 

Thanks to manrajsingh for suggesting above solution.

This article is contributed by Abhay. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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