Count of distinct integers in range [1, N] that do not have any subset sum as K

Given two positive integers N and K such that K?N, the task is to find the maximum number of distinct integers in the range [1, N] having no subset with a sum equal to K. If there are multiple solutions, print any.

Examples:

Input: N = 5, K = 3
Output: 1 4 5
Explanation: There are two sets of distinct numbers of size 3 which don't have any subset sum of 3.
These are {1, 4, 5} and {2, 4, 5}. So, print any of them in any order.

Input: N = 1, K =1
Output: 0

Approach: The idea is based on the following observations:

• Any number greater than K can be chosen as they can never contribute to a subset whose sum is K.
• K cannot be chosen.
• For the numbers less than K, at most K/2 numbers can be chosen. For example:
  • If K=5, 1+4=5, and 2+3=5, so either 1 can be chosen or 4 and similarly either 2 or 3 can be chosen. Thus, at most (5/2=2) numbers can be chosen.
  • If K=6, 1+5=6, 2+4=6 and 3+3=6. Again, 3 numbers can be chosen such that no subset-sum equals 6. 3 can always be chosen as only distinct numbers are being chosen, and either 1 or 5 and similarly either 2 or 4 can be chosen. Thus, at most (6/3=3) numbers can be chosen.
• Therefore, the maximum number of distinct numbers that can be chosen is (N-K)+(K/2).

Follow the below steps to solve the problem:

• The maximum number of distinct digits that can be chosen is (N-K)+(K/2).
• All the numbers greater than K need to be chosen i.e N-K numbers from the end. K/2 elements less than K also need to be chosen.
• Thus, a possible solution is to choose (N-K)+(K/2) consecutive numbers starting from N and excluding K.
• The easiest way to do this is to create an array storing all values from 1 to N, except for K, reverse the array, and print (N-K)+(K/2) elements from the beginning.

Below is the implementation of the above approach:

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

// Function to find maximum number of distinct integers
// in [1, N] having no subset with sum equal to K
void findSet(int N, int K)
{
    // Declare a vector to store
    // the required numbers
    vector<int> a;

    // Store all the numbers in [1, N] except K
    for (int i = 1; i <= N; i++) {
        if (i != K)
            a.push_back(i);
    }

    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);

    // Reverse the array
    reverse(a.begin(), a.end());

    // Print the required numbers
    for (int i = 0; i < MaxDistinct; i++)
        cout << a[i] << " ";
}

// Driver Code
int main()
{
    // Given Input
    int N = 5, K = 3;

    // Function Call
    findSet(N, K);

    return 0;
}

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find maximum number of distinct 
// integers in [1, N] having no subset with 
// sum equal to K
static void findSet(int N, int K)
{
    
    // Declare a vector to store
    // the required numbers
    ArrayList<Integer> a = new ArrayList<Integer>();

    // Store all the numbers in [1, N] except K
    for(int i = 1; i <= N; i++)
    {
        if (i != K)
            a.add(i);
    }

    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);

    // Reverse the array
    Collections.reverse(a);

    // Print the required numbers
    for(int i = 0; i < MaxDistinct; i++)
        System.out.print(a.get(i) + " ");
}

// Driver Code
public static void main(String[] args)
{
    
    // Given Input
    int N = 5, K = 3;
    
    // Function Call
    findSet(N, K);
}
}

// This code is contributed by sanjoy_62

# Python3 program for the above approach

# Function to find maximum number of distinct
# integers in [1, N] having no subset with
# sum equal to K
def findSet(N, K):
    
    # Declare a vector to store
    # the required numbers
    a = []

    # Store all the numbers in [1, N] except K
    for i in range(1, N + 1):
        if (i != K):
            a.append(i)

    # Store the maximum number
    # of distinct numbers
    MaxDistinct = (N - K) + (K // 2)

    # Reverse the array
    a = a[::-1]
    
    # Print the required numbers
    for i in range(MaxDistinct):
        print(a[i], end = " ")

# Driver Code
if __name__ == '__main__':
    
    # Given Input
    N = 5
    K = 3

    # Function Call
    findSet(N, K)

# This code is contributed by mohit kumar 29

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

class GFG{

// Function to find maximum number of distinct 
// integers in [1, N] having no subset with 
// sum equal to K
static void findSet(int N, int K)
{
    
    // Declare a vector to store
    // the required numbers
    List<int> a = new List<int>();

    // Store all the numbers in [1, N] except K
    for(int i = 1; i <= N; i++)
    {
        if (i != K)
            a.Add(i);
    }

    // Store the maximum number
    // of distinct numbers
    int MaxDistinct = (N - K) + (K / 2);

    // Reverse the array
    a.Reverse();

    // Print the required numbers
    for(int i = 0; i < MaxDistinct; i++)
        Console.Write(a[i] + " ");
}

// Driver Code
public static void Main()
{
    // Given Input
    int N = 5, K = 3;
    
    // Function Call
    findSet(N, K);
}
}

// This code is contributed by avijitmondal1998.

  <script>
  
// JavaScript program for the above approach

// Function to find maximum number of distinct integers
// in [1, N] having no subset with sum equal to K
function findSet( N,  K)
{
    // Declare a vector to store
    // the required numbers
    let a = [];

    // Store all the numbers in [1, N] except K
    for (let i = 1; i <= N; i++) {
        if (i != K)
            a.push(i);
    }

    // Store the maximum number
    // of distinct numbers
    let MaxDistinct = (N - K) + parseInt(K / 2);

    // Reverse the array
    a.reverse();
    // Print the required numbers
    for (let i = 0; i < MaxDistinct; i++)
        document.write(a[i]+" ");
}

// Driver Code
// Given Input
    let N = 5, K = 3;

    // Function Call
    findSet(N, K);

  // This code is contributed by Potta Lokesh
  
</script>

5 4 2 

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