Find Kth number that can be written as sum of different powers of N

Given two positive integers N and K. The task is to find the Kth number that can be written as the sum of different non-negative powers of N. 

Examples:

Input: N = 3, K = 4
Output: 9
Explanation:
First number that can be written as sum of powers of 3 is [1 = 3⁰]
Second number that can be written as sum of powers of 3 is [3 = 3¹]
Third number that can be written as sum of powers of 3 is [4 = 3⁰ + 3¹]
Fourth number that can be written as sum of powers of 3 is [9 = 3²]
Therefore answer is 9.

Input: N = 2, K = 12
Output: 12

Approach: This problem can be solved by using the concept of decimal to binary conversion.  The idea is to find the binary representation of K and start iterating from the least significant bit to the most significant bit. If the current bit is set then include the corresponding power of N to the answer otherwise skip that bit. Refer to the picture below for a better understanding.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find Kth number that can be
// represented as sum of different
// non-negative powers of N
int FindKthNum(int n, int k)
{
    // To store the ans
    int ans = 0;

    // Value of current power of N
    int currPowOfN = 1;

    // Iterating through bits of K
    // from LSB to MSB
    while (k) {
        // If the current bit is 1 then include
        // corresponding power of n into ans
        if (k & 1) {
            ans = ans + currPowOfN;
        }

        currPowOfN = currPowOfN * n;
        k = k / 2;
    }

    // Return the result
    return ans;
}

// Driver Code
int main()
{
    int N = 3;
    int K = 4;

    cout << FindKthNum(N, K);
}

// Java program for the above approach
class GFG
{
  
  // Function to find Kth number that can be
  // represented as sum of different
  // non-negative powers of N
  static int FindKthNum(int n, int k)
  {
    
      // To store the ans
      int ans = 0;

      // Value of current power of N
      int currPowOfN = 1;

      // Iterating through bits of K
      // from LSB to MSB
      while (k > 0)
      {
        
          // If the current bit is 1 then include
          // corresponding power of n into ans
          if ((k & 1) == 1) {
              ans = ans + currPowOfN;
          }

          currPowOfN = currPowOfN * n;
          k = k / 2;
      }

      // Return the result
      return ans;
  }
 
  // Driver Code
  public static void main(String []args)
  {
      int N = 3;
      int K = 4;

      System.out.println(FindKthNum(N, K));
  }

}

// This Code is contributed by ihritik

# Python program for the above approach

# Function to find Kth number that can be
# represented as sum of different
# non-negative powers of N
def FindKthNum(n, k):

  # To store the ans
  ans = 0

  # value of current power of N
  currPowOfN = 1

  # Iterating through bits of K
  # from LSB to MSB
  while (k > 0):
    
    # If the current bit is 1 then include
    # corresponding power of n into ans
    if ((k & 1) == 1) :
      ans = ans + currPowOfN
    

    currPowOfN = currPowOfN * n
    k = k // 2
  

  # Return the result
  return ans



# Driver Code

N = 3
K = 4
print(FindKthNum(N, K));
  

# This Code is contributed by ihritik

// C# program for the above approach
using System;
class GFG
{
  
  // Function to find Kth number that can be
  // represented as sum of different
  // non-negative powers of N
  static int FindKthNum(int n, int k)
  {
    
      // To store the ans
      int ans = 0;

      // Value of current power of N
      int currPowOfN = 1;

      // Iterating through bits of K
      // from LSB to MSB
      while (k > 0) 
      {
        
          // If the current bit is 1 then include
          // corresponding power of n into ans
          if ((k & 1) == 1) {
              ans = ans + currPowOfN;
          }

          currPowOfN = currPowOfN * n;
          k = k / 2;
      }

      // Return the result
      return ans;
  }
 

  // Driver Code
  public static void Main()
  {
      int N = 3;
      int K = 4;

      Console.WriteLine(FindKthNum(N, K));
  }

}

// This Code is contributed by ihritik

<script>
// JavaScript program for the above approach

// Function to find Kth number that can be
// represented as sum of different
// non-negative powers of N
function FindKthNum(n, k)
{

    // To store the ans
    let ans = 0;

    // Value of current power of N
    let currPowOfN = 1;

    // Iterating through bits of K
    // from LSB to MSB
    while (k)
    {
    
        // If the current bit is 1 then include
        // corresponding power of n into ans
        if (k & 1)
        {
            ans = ans + currPowOfN;
        }

        currPowOfN = currPowOfN * n;
        k = Math.floor(k / 2);
    }

    // Return the result
    return ans;
}

// Driver Code
let N = 3;
let K = 4;
document.write(FindKthNum(N, K));

// This code is contributed by rohitsingh07052.
</script>

9

Time Complexity: O(log₂K)

Space Complexity: O(1)