Given two integers N and K, the task is to find the Nth 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 Nth 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 Nth 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:
- p = log2(6) = 2
- 32 = 9, Subset = {9}
- N = 6 % 4 = 2
2nd iteration:
- p = log2(2) = 1
- 31 = 3, Subset = {3, 9}
- 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 Kp. Insert it into the front of the subset.
- Update N to N % 2p.
- 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 3count 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)