Given an integer N and a set of all non-negative powers of N as S = {N0, N1, N2, N3, … }, arrange all non-empty subsets of S in increasing order of subset-sum. The task is to find the sum of the greatest and smallest elements of the Kth subset from that ordering.
Examples:
Input: N = 4, K = 3
Output: 5
Explanation:
S = {1, 4, 16, 64, …}.
Therefore, the non-empty subsets arranged in increasing order of their sum = {{1}, {4}, {1, 4}, {16}, {1, 16}, {4, 16}, {1, 4, 16}………}.
So the elements of the subset in Kth(3rd) subset are {1, 4}. So the sum of the greatest and smallest elements of this subset = (4 + 1) = 5.Input: N = 3, K = 4
Output: 18
Explanation:
S = {1, 3, 9, 27, 81, …}.
Therefore, the non-empty subsets arranged in increasing order of their sum = {{1}, {3}, {1, 3}, {9}, {1, 9}, {3, 9}, {1, 3, 9} ……..}.
So the element in the subset at 4th position is {9}. So the sum of the greatest and smallest elements of this subset = (9 + 9) = 18.
Approach: This approach is based on the concept of Power-set. Follow the steps below to solve the problem:
- Generate the corresponding binary expression of the integer K (i.e., the position of the subset) in inverse order to maintain the increasing sum of elements in the subset.
- Then calculate whether there will be an element in the subset at corresponding positions depending on the bits present in lst[] list at successive positions.
- Iterate over the lst[] and if lst[i] is 0, then ignore that bit, otherwise (Ni)*lst[i] will be in the Kth Subset num[].
- Then the sum is calculated by taking the element of the num[] at position 0 and at the last position.
- Print the resultant sum after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of greatest// and smallest element of Kth Subsetvoid sumofExtremes(int N, int K){ // Stores the binary equivalent // of k in inverted order list<int> lst; while (K > 0) { lst.push_back(K % 2); K = K / 2; } // Stores the kth subset list<int> num; int x = 0; // Iterate over the list for(auto element : lst) { // If the element is non-zero if (element != 0) { int a = pow(N, x); a = a * (element); num.push_back(a); } x++; } // Update S to length of num int s = num.size(); // If length of the subset is 1 if (s == 1) // Print twice of that element cout << (2 * num.front()) << "\n"; // Print the sum of first and // last element else cout << num.front() + num.back();}// Driver Codeint main(){ // Given number N int N = 4; // Given position K int K = 6; sumofExtremes(N, K);}// This code is contributed by akhilsaini |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{// Function to find the sum of greatest// and smallest element of Kth Subsetpublic static void sumofExtremes(int N, int K){ // Stores the binary equivalent // of k in inverted order List<Integer> lst = new ArrayList<Integer>(); while (K > 0) { lst.add(K % 2); K = K / 2; } // Stores the kth subset List<Integer> num = new ArrayList<Integer>(); int x = 0; // Iterate over the list for(int element : lst) { // If the element is non-zero if (element != 0) { int a = (int)Math.pow(N, x); a = a * (element); num.add(a); } x++; } // Update S to length of num int s = num.size(); // If length of the subset is 1 if (s == 1) // Print twice of that element System.out.println(2 * num.get(0)); // Print the sum of first and // last element else System.out.println(num.get(0) + num.get(s - 1));}// Driver Codepublic static void main(String[] args){ // Given number N int N = 4; // Given position K int K = 6; // Function call sumofExtremes(N, K);}}// This code is contributed by akhilsaini |
Python3
# Python3 program for the above approach# Function to find the sum of greatest# and smallest element of Kth Subsetdef sumofExtremes(N, K): # Stores the binary equivalent # of k in inverted order lst = [] while K > 0: lst.append(K % 2) K = K//2 # Stores the kth subset num = [] # Iterate over the list for i in range(0, len(lst)): # If the element is non-zero if(lst[i] != 0): a = N**i a = a * lst[i] num.append(a) # Update S to length of num s = len(num) # If length of the subset is 1 if(s == 1): # Print twice of that element print(2 * num[0]) # Print the sum of first and # last element else: print(num[0] + num[s - 1])# Driver Code# Given number NN = 4# Given position KK = 6# Function CallsumofExtremes(N, K) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the sum of greatest// and smallest element of Kth Subsetpublic static void sumofExtremes(int N, int K){ // Stores the binary equivalent // of k in inverted order List<int> lst = new List<int>(); while (K > 0) { lst.Add(K % 2); K = K / 2; } // Stores the kth subset List<int> num = new List<int>(); // Iterate over the list for(int i = 0; i < lst.Count; i++) { // If the element is non-zero if (lst[i] != 0) { int a = (int)Math.Pow(N, i); a = a * (lst[i]); num.Add(a); } } // Update S to length of num int s = num.Count; // If length of the subset is 1 if (s == 1) // Print twice of that element Console.WriteLine(2 * num[0]); // Print the sum of first and // last element else Console.WriteLine(num[0] + num[s - 1]);}// Driver Codestatic public void Main(){ // Given number N int N = 4; // Given position K int K = 6; // Function call sumofExtremes(N, K);}}// This code is contributed by akhilsaini |
Javascript
<script>// Javascript program for the above approach// Function to find the sum of greatest// and smallest element of Kth Subsetfunction sumofExtremes(N, K){ // Stores the binary equivalent // of k in inverted order let lst = []; while (K > 0) { lst.push(K % 2); K = Math.floor(K / 2); } // Stores the kth subset let num = []; let x = 0; // Iterate over the list for(let element of lst) { // If the element is non-zero if (element != 0) { let a = Math.pow(N, x); a = a * (element); num.push(a); } x++; } // Update S to length of num let s = num.length; // If length of the subset is 1 if (s == 1) // Print twice of that element document.write((2 * num[0]) + "+"); // Print the sum of first and // last element else document.write(num[0] + num[num.length - 1]);}// Driver Code // Given number N let N = 4; // Given position K let K = 6; sumofExtremes(N, K);// This code is contributed by gfgking</script> |
Output:
20
Time Complexity: O(log K)
Auxiliary Space: O(N)
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