Count of subsets of integers from 1 to N having no adjacent elements

Given an integer N, the task is to count the number of subsets formed from an array of integers from 1 to N which doesn’t contain adjacent elements. A subset can’t be chosen if it satisfies the non-adjacent element condition, but it is possible to add more elements.
Examples:

Input: N = 4
Output: 3
Explanation:
Array is {1, 2, 3, 4}
So to satisfy the condition, the subsets formed are :
{1, 3}, {2, 4}, {1, 4}

Input: N = 5
Output: 4

Approach:
This problem can be solved by using Dynamic Programming. For the last element, we have two choices, either we include it, or we exclude it. Let DP[i] be the number of our desirable subsets ending at index i. 
 So next Subproblem becomes DP[i-3]
So the DP relation becomes : 

DP[i] = DP[i-2] + DP[i-3]  

But, we need to observe the base cases: 

• When N=0, we cannot form any subset with 0 numbers.
• When N=1, we can form 1 subset, {1}
• When N=2, we can form 2 subsets, {1}, {2}
• When N=3, we can form 2 subsets, {1, 3}, {2}

Below is the implementation of above approach: 

C++

// C++ Code to count subsets not containing
// adjacent elements from 1 to N

#include <bits/stdc++.h>
using namespace std;

// Function to count subsets
int countSubsets(int N)
{

    if(N <= 2)
        return N;

    if(N == 3)
        return 2;

    int DP[N + 1] = {0};

    DP[0] = 0, DP[1] = 1, DP[2] = 2, DP[3] = 2;

    for (int i = 4; i <= N; i++) {

        DP[i] = DP[i - 2] + DP[i - 3];
    }

    return DP[N];
}

// Driver Code
int main()
{
    int N = 20;

    cout << countSubsets(N);

    return 0;
}

Java

// Java code to count subsets not containing
// adjacent elements from 1 to N
class GFG{

// Function to count subsets
static int countSubsets(int N)
{
    if(N <= 2)
       return N;

    if(N == 3)
       return 2;

    int []DP = new int[N + 1];

    DP[0] = 0;
    DP[1] = 1;
    DP[2] = 2;
    DP[3] = 2;

    for(int i = 4; i <= N; i++)
    {
       DP[i] = DP[i - 2] + DP[i - 3];
    }
    return DP[N];
}

// Driver code
public static void main(String[] args)
{
    int N = 20;

    System.out.print(countSubsets(N));
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 Code to count subsets
# not containing adjacent elements
# from 1 to N

# Function to count subsets
def countSubsets(N):

    if(N <= 2):
        return N

    if(N == 3):
        return 2

    DP = [0] * (N + 1)

    DP[0] = 0
    DP[1] = 1
    DP[2] = 2
    DP[3] = 2

    for i in range(4, N + 1):

        DP[i] = DP[i - 2] + DP[i - 3]

    return DP[N]

# Driver Code
if __name__ == '__main__':
    N = 20

    print(countSubsets(N))

# This code is contributed by Mohit Kumar

C#

// C# code to count subsets not containing
// adjacent elements from 1 to N
using System;
class GFG{

// Function to count subsets
static int countSubsets(int N)
{
    if(N <= 2)
        return N;

    if(N == 3)
        return 2;

    int []DP = new int[N + 1];

    DP[0] = 0;
    DP[1] = 1;
    DP[2] = 2;
    DP[3] = 2;

    for(int i = 4; i <= N; i++)
    {
        DP[i] = DP[i - 2] + DP[i - 3];
    }
    return DP[N];
}

// Driver code
public static void Main(String[] args)
{
    int N = 20;

    Console.Write(countSubsets(N));
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>
// javascript code to count subsets not containing
// adjacent elements from 1 to N
// Function to count subsets
    function countSubsets(N) {
        if (N <= 2)
            return N;

        if (N == 3)
            return 2;

        var DP = Array(N + 1).fill(0);

        DP[0] = 0;
        DP[1] = 1;
        DP[2] = 2;
        DP[3] = 2;

        for (i = 4; i <= N; i++) {
            DP[i] = DP[i - 2] + DP[i - 3];
        }
        return DP[N];
    }

    // Driver code

        var N = 20;

        document.write(countSubsets(N));

// This code contributed by gauravrajput1
</script>

Output:

265

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

Efficient approach: Space-optimized approach

In this approach, we optimize the space by using variables.

Implementation steps :

• Initialize variables to store previous computations : prev1, prev2 , prev3, prev4 
• The current computation is depend upon the prev3 and prev2.
• After every iteration update these variables to iterate further.
• At last return answer stored in curr.

Implementation:

C++

// C++ Code to count subsets not containing
// adjacent elements from 1 to N

#include <bits/stdc++.h>
using namespace std;

// Function to count subsets
int countSubsets(int N)
{


    // Base Case
    if(N <= 2)
        return N;

    if(N == 3)
        return 2;


    int prev1 = 0, prev2 = 1, prev3 = 2, prev4 = 2 , curr;

    for (int i = 4; i <= N; i++) {

        curr = prev3 + prev2;

        // assigning value to iterate further
        prev1 = prev2;
        prev2= prev3;
        prev3 = prev4;
        prev4 = curr;
     }
    // return answer
    return curr;
}

// Driver Code
int main()
{
    int N = 20;

    // function call
    cout << countSubsets(N);

    return 0;
}

Java

// Java program for above approach

import java.util.*;

public class Main {

    // Function to count subsets
    public static int countSubsets(int N)
    {

        // Base Case
        if (N <= 2) {
            return N;
        }

        if (N == 3) {
            return 2;
        }

        int prev1 = 0, prev2 = 1, prev3 = 2;
        int prev4 = 2, curr = 0;

        // loop to get the current value
        // from previous computations
        for (int i = 4; i <= N; i++) {
            curr = prev3 + prev2;

            // assigning value to iterate further
            prev1 = prev2;
            prev2 = prev3;
            prev3 = prev4;
            prev4 = curr;
        }

        // return answer
        return curr;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 20;
        System.out.println(countSubsets(N));
    }
}

Python3

# Python3 Code to count subsets not containing
# adjacent elements from 1 to N

# Function to count subsets
def countSubsets(N):

    # Base Case
    if(N <= 2):
        return N

    if(N == 3):
        return 2

    prev1 = 0
    prev2 = 1
    prev3 = 2
    prev4 = 2
    curr = 0

    for i in range(4, N+1):
        curr = prev3 + prev2

        # assigning value to iterate further
        prev1 = prev2
        prev2 = prev3
        prev3 = prev4
        prev4 = curr

    # return answer
    return curr


# Driver Code
N = 20

# function call
print(countSubsets(N))

C#

// C# program for above approach

using System;

class MainClass {
    // Function to count subsets
    public static int CountSubsets(int N)
    {

        // Base Case
        if (N <= 2) {
            return N;
        }

        if (N == 3) {
            return 2;
        }

        int prev1 = 0, prev2 = 1, prev3 = 2, prev4 = 2,
            curr = 0;

        // loop to get the current value from previous
        // computations
        for (int i = 4; i <= N; i++) {
            curr = prev3 + prev2;

            // assigning value to iterate further
            prev1 = prev2;
            prev2 = prev3;
            prev3 = prev4;
            prev4 = curr;
        }

        // return answer
        return curr;
    }

    // Driver Code
    public static void Main(string[] args)
    {

        int N = 20;

        // function call
        Console.WriteLine(CountSubsets(N));
    }
}

Javascript

// Function to count subsets
function countSubsets(N) {
    // Base Case
    if (N <= 2) {
        return N;
    }
    if (N == 3) {
        return 2;
    }
    let prev1 = 0, prev2 = 1, prev3 = 2, prev4 = 2 , curr;
    for (let i = 4; i <= N; i++) {
        curr = prev3 + prev2;
        // assigning value to iterate further
        prev1 = prev2;
        prev2= prev3;
        prev3 = prev4;
        prev4 = curr;
     }
    // return answer
    return curr;
}

// Driver Code
const N = 20;

// function call
console.log(countSubsets(N));

Output

265

Time Complexity: O(N)
Auxiliary Space: O(1)