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)