Nth Subset of the Sequence consisting of powers of K in increasing order of their Sum

Given two integers N and K, the task is to find the N^th Subset from the sequence of subsets generated from the powers of K i.e. {1, K¹, K², K³, .....} such that the subsets are arranged in increasing order of their sum, the task is to find the N^th subset from the sequence.

Examples:

Input: N = 5, K = 3 
Output: 1 9 
Explanation: 
The sequence of subsets along with their sum are:

• Subset = {1}, Sum = 1
• Subset = {3}, Sum = 3
• Subset = {1, 3}, Sum = 4
• Subset = {9}, Sum = 9
• Subset = {1, 9}, Sum = 10

Therefore, the subset at position 5 is {1, 9}.

Input: N = 4, K = 4 
Output: 16

Approach: 
Let's refer to the required sequence for K = 3 given below:

From the above sequence, it can be observed that the subset {3} has position 2, the subset {9} has position 4, and the subset {27} has position 8, and so on. The subset {1, 3}, {1, 9}, {1, 27} occupies positions 3, 5, and 9 respectively. Hence, all the elements of the required N^th subset can be obtained by finding the nearest power of 2 which is smaller than or equal to N.

Illustration: 
N = 6, K = 3
1st iteration:

1. p = log₂(6) = 2
2. 3² = 9, Subset = {9}
3. N = 6 % 4 = 2

2nd iteration:

1. p = log₂(2) = 1
2. 3¹ = 3, Subset = {3, 9}
3. N = 2 % 2 = 0

Therefore, the required subset is {3, 9}

Follow the steps below to solve the problem:

• Calculate the nearest power of 2 which is smaller than or equal to N, say p. Therefore, p = log₂N.
• Now, the element of the subset will be K^p. Insert it into the front of the subset.
• Update N to N % 2^p.
• Repeat the above steps until N becomes 0, and consequently print the obtained subset.

Below is the implementation of the above approach:

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

// Function to print the 
// required N-th subset 
void printSubset(lli n, int k)
{
    vector<lli> answer;
    while(n > 0)
    {

        // Nearest power of 2<=N
        lli p = log2(n);
        
        // Now insert k^p in the answer
        answer.push_back(pow(k, p));
        
        // update n
        n %= (int)pow(2, p);
    }

    // Print the ans in sorted order
    reverse(answer.begin(), answer.end());
    for(auto x: answer)
    {
        cout << x << " ";
    }
}

// Driver Code
int main()
{
    lli n = 5;
    int k = 4;
    printSubset(n, k);
}

// This code is contributed by winter_soldier

// Java program for above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{

  // Function to print the  
  // required N-th subset  
  static void printSubset(long n, int k) 
  { 
    ArrayList<Long> answer = new ArrayList<>(); 
    while(n > 0) 
    { 

      // Nearest power of 2<=N 
      long p = (long)(Math.log(n) / Math.log(2));; 

      // Now insert k^p in the answer 
      answer.add((long)(Math.pow(k, p))); 

      // update n 
      n %= (int)Math.pow(2, p); 
    } 

    // Print the ans in sorted order 
    Collections.sort(answer);
    for(Long x: answer) 
    { 
      System.out.print(x + " "); 
    } 
  }

  // Driver function
  public static void main (String[] args)
  {
    long n = 5; 
    int k = 4; 
    printSubset(n, k);
  }
}

// This code is contributed by offbeat

# Python3 program for
# the above approach
import math

# Function to print the
# required N-th subset
def printSubset(N, K):
    # Stores the subset
    answer = ""
    while(N > 0):
        # Nearest power of 2 <= N
        p = int(math.log(N, 2))
        # Insert K ^ p in the subset
        answer = str(K**p)+" "+answer
        # Update N
        N = N % (2**p)
        
    # Print the subset
    print(answer)
    
# Driver Code
N = 5
K = 4
printSubset(N, K)

// C# program for the above approach 
using System;
using System.Collections.Generic;
class GFG {

  // Function to print the 
  // required N-th subset 
  static void printSubset(int n, int k)
  {
    List<int> answer = new List<int>();
    while(n > 0)
    {

      // Nearest power of 2<=N
      int p = (int)Math.Log(n,2);

      // Now insert k^p in the answer
      answer.Add((int)Math.Pow(k, p));

      // update n
      n %= (int)Math.Pow(2, p);
    }

    // Print the ans in sorted order
    answer.Reverse();
    foreach(int x in answer)
    {
      Console.Write(x + " ");
    }
  }

  // Driver code
  static void Main() {
    int n = 5;
    int k = 4;
    printSubset(n, k);
  }
}

// This code is contributed by divyeshrabadiya07.

<script>
// Javascript program for the above approach 

// Function to print the 
// required N-th subset 
function printSubset(n, k)
{
    var answer = [];
    while(n > 0)
    {

        // Nearest power of 2<=N
        var p = parseInt(Math.log2(n));
        
        // Now insert k^p in the answer
        answer.push(Math.pow(k, p));
        
        // update n
        n %= parseInt(Math.pow(2, p));
    }

    // Print the ans in sorted order
    answer.sort();
    //reverse(answer.begin(), answer.end());
    for(var i=0;i<answer.length;i++)
    {
        document.write(answer[i] + " ");
    }
}

var n = 5;
var k = 4;
printSubset(n, k);

//This code is contributed by SoumikMondal
</script>

1 16 

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

Approach: 

• Initialize the count and x by 0. Also, a vector to store the elements of the subsets.
• Do the following while n is greater than 0.
  • Set x = n & 1, for finding if the last bit of the number is set or not.
  • Now Push element 3^count into the subset if n is not 0.
  • Reduce the number n by two with the help of right shifting by 1 unit.
  • Increase the count value by 1.
• Finally, the elements in the array are the elements of the Nth subset.

Below is the implementation of the above approach:

// C++ program to print subset
// at the nth position ordered
// by the sum of the elements
#include <bits/stdc++.h>
using namespace std;

// Function to print the elements of 
// the subset at pos n
void printsubset(int n,int k)
{
    //  Initialize count=0 and x=0
    int count = 0, x = 0;
  
    // create a vector for 
    // storing the elements
    // of subsets
    vector<int> vec;
  
    // doing until all the
    // set bits of n are used
    while (n) {
        x = n & 1;
      
        // this part is executed only
        // when the last bit is
        // set
        if (x) {
            vec.push_back(pow(k, count));
        }
      
        // right shift the bit by one position
        n = n >> 1; 
      
        // increasing the count each time by one
        count++; 
    }
  
    // printing the values os elements
    for (int i = 0; i < vec.size(); i++)
        cout << vec[i] << " ";
}

// Driver Code
int main()
{
    int n = 7,k=4;
    printsubset(n,k);
    return 0;
}

// This code is contributed by shivkant

// Java program to print subset
// at the nth position ordered
// by the sum of the elements
import java.util.*;
import java.lang.*;
class GFG{
  
// Function to print the 
// elements of the subset 
// at pos n
static void printsubset(int n, 
                        int k)
{
  // Initialize count=0 and x=0
  int count = 0, x = 0;

  // Create a vector for 
  // storing the elements
  // of subsets
  ArrayList<Integer> vec = 
            new ArrayList<>();

  // Doing until all the
  // set bits of n are used
  while (n != 0) 
  {
    x = n & 1;

    // This part is executed only
    // when the last bit is
    // set
    if (x != 0) 
    {
      vec.add((int)Math.pow(k, 
                            count));
    }

    // Right shift the bit 
    // by one position
    n = n >> 1; 

    // Increasing the count 
    // each time by one
    count++; 
  }

  // Printing the values os elements
  for (int i = 0; i < vec.size(); i++)
    System.out.print(vec.get(i) + " ");
}

// Driver function
public static void main (String[] args) 
{
  int n = 7, k = 4;
  printsubset(n, k);
}
}

// This code is contributed by offbeat

# Python3 program to print subset
# at the nth position ordered
# by the sum of the elements
import math

# Function to print the elements of 
# the subset at pos n
def printsubset(n, k):
    
    # Initialize count=0 and x=0
    count = 0
    x = 0

    # Create a vector for 
    # storing the elements
    # of subsets
    vec = []

    # Doing until all the
    # set bits of n are used
    while (n > 0):
        x = n & 1
        
        # This part is executed only
        # when the last bit is
        # set
        if (x):
            vec.append(pow(k, count))
    
        # Right shift the bit by one position
        n = n >> 1
    
        # Increasing the count each time by one
        count += 1 

    # Printing the values os elements
    for item in vec:
        print(item, end = " ")

# Driver Code
n = 7
k = 4

printsubset(n, k)

# This code is contributed by Stream_Cipher

// C# program to print subset
// at the nth position ordered
// by the sum of the elements
using System.Collections.Generic; 
using System;

class GFG{

// Function to print the 
// elements of the subset 
// at pos n
static void printsubset(int n, int k)
{
    
    // Initialize count=0 and x=0
    int count = 0, x = 0;
    
    // Create a vector for 
    // storing the elements
    // of subsets
    List<int> vec = new List<int>();
    
    // Doing until all the
    // set bits of n are used
    while (n != 0) 
    {
        x = n & 1;
    
        // This part is executed only
        // when the last bit is
        // set
        if (x != 0) 
        {
            vec.Add((int)Math.Pow(k, count));
        }
    
        // Right shift the bit 
        // by one position
        n = n >> 1; 
    
        // Increasing the count 
        // each time by one
        count++; 
    }
    
    // Printing the values os elements
    for(int i = 0; i < vec.Count; i++)
        Console.Write(vec[i] + " ");
}

// Driver code
public static void Main () 
{
    int n = 7, k = 4;
    
    printsubset(n, k);
}
}

// This code is contributed by Stream_Cipher

<script>
    // Javascript program to print subset
    // at the nth position ordered
    // by the sum of the elements
    
    // Function to print the
    // elements of the subset
    // at pos n
    function printsubset(n, k)
    {

        // Initialize count=0 and x=0
        let count = 0, x = 0;

        // Create a vector for
        // storing the elements
        // of subsets
        let vec = [];

        // Doing until all the
        // set bits of n are used
        while (n != 0)
        {
            x = n & 1;

            // This part is executed only
            // when the last bit is
            // set
            if (x != 0)
            {
                vec.push(Math.pow(k, count));
            }

            // Right shift the bit
            // by one position
            n = n >> 1;

            // Increasing the count
            // each time by one
            count++;
        }

        // Printing the values os elements
        for(let i = 0; i < vec.length; i++)
            document.write(vec[i] + " ");
    }
    
    let n = 7, k = 4;
     
    printsubset(n, k);

</script>

1 4 16 

Time Complexity: O(log₂N) 
Auxiliary Space: O(log₂N)