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# Sum of maximum and minimum of Kth subset ordered by increasing subset sum

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:

1. 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.
2. 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.
3. 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[].
4. Then the sum is calculated by taking the element of the num[] at position 0 and at the last position.
5. Print the resultant sum after the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the sum of greatest``// and smallest element of Kth Subset``void` `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 Code``int` `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 approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG{` `// Function to find the sum of greatest``// and smallest element of Kth Subset``public` `static` `void` `sumofExtremes(``int` `N, ``int` `K)``{``    ` `    ``// Stores the binary equivalent``    ``// of k in inverted order``    ``List lst = ``new` `ArrayList();` `    ``while` `(K > ``0``)``    ``{``        ``lst.add(K % ``2``);``        ``K = K / ``2``;``    ``}` `    ``// Stores the kth subset``    ``List num = ``new` `ArrayList();` `    ``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 Code``public` `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 Subset``def` `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 N``N ``=` `4` `# Given position K``K ``=` `6` `# Function Call``sumofExtremes(N, K)`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the sum of greatest``// and smallest element of Kth Subset``public` `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);` `    ``// Print the sum of first and``    ``// last element``    ``else``        ``Console.WriteLine(num + num[s - 1]);``}` `// Driver Code``static` `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

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Output:

`20`

Time Complexity: O(log K)
Auxiliary Space: O(N)