Count non-adjacent subsets from numbers arranged in Circular fashion

Given that N people are sitting in a circular queue numbered from 1 to N, the task is to count the number of ways to select a subset of them such that no two consecutive are sitting together. The answer could be large so compute the answer modulo 109 + 7. Note that an empty subset is also a valid subset.

Examples:

Input: N = 2
Output: 3
All possible subsets are {}, {1} and {2}.



Input: N = 3
Output: 4

Approach: Let’s find the answer for small values of N.

N = 1 -> All possible subset are {}, {1}.
N = 2 -> All possible subset are {}, {1}, {2}.
N = 3 -> All possible subset are {}, {1}, {2}, {3}.
N = 4 -> All possible subset are {}, {1}, {2}, {3}, {4}, {1, 3}, {2, 4}.
So the sequence will be 2, 3, 4, 7, …
When N = 5 the count will be 11 and if N = 6 then the count will be 18.
It can now be observed that the sequence is similar to a Fibonacci series starting from the second term with the first two terms as 3 and 4.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
  
const ll N = 10000;
const ll MOD = 1000000007;
  
ll F[N];
  
// Function to pre-compute the sequence
void precompute()
{
  
    // For N = 1 the answer will be 2
    F[1] = 2;
  
    // Starting two terms of the sequence
    F[2] = 3;
    F[3] = 4;
  
    // Comute the rest of the sequence
    // with the relation
    // F[i] = F[i - 1] + F[i - 2]
    for (int i = 4; i < N; i++)
        F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
  
// Driver code
int main()
{
    int n = 8;
  
    // Pre-compute the sequence
    precompute();
  
    cout << F[n];
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG 
{
static int N = 10000;
static int MOD = 1000000007;
  
static int []F = new int[N];
  
// Function to pre-compute the sequence
static void precompute()
{
  
    // For N = 1 the answer will be 2
    F[1] = 2;
  
    // Starting two terms of the sequence
    F[2] = 3;
    F[3] = 4;
  
    // Comute the rest of the sequence
    // with the relation
    // F[i] = F[i - 1] + F[i - 2]
    for (int i = 4; i < N; i++)
        F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
  
// Driver code
public static void main(String []args)
{
    int n = 8;
  
    // Pre-compute the sequence
    precompute();
  
    System.out.println(F[n]);
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python implementation of the approach
N = 10000;
MOD = 1000000007;
  
F = [0] * N;
  
# Function to pre-compute the sequence
def precompute():
  
    # For N = 1 the answer will be 2
    F[1] = 2;
  
    # Starting two terms of the sequence
    F[2] = 3;
    F[3] = 4;
  
    # Comute the rest of the sequence
    # with the relation
    # F[i] = F[i - 1] + F[i - 2]
    for i in range(4,N):
        F[i] = (F[i - 1] + F[i - 2]) % MOD;
  
# Driver code
n = 8;
  
# Pre-compute the sequence
precompute();
print(F[n]);
  
# This code is contributed by 29AjayKumar

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
static int N = 10000;
static int MOD = 1000000007;
  
static int []F = new int[N];
  
// Function to pre-compute the sequence
static void precompute()
{
  
    // For N = 1 the answer will be 2
    F[1] = 2;
  
    // Starting two terms of the sequence
    F[2] = 3;
    F[3] = 4;
  
    // Comute the rest of the sequence
    // with the relation
    // F[i] = F[i - 1] + F[i - 2]
    for (int i = 4; i < N; i++)
        F[i] = (F[i - 1] + F[i - 2]) % MOD;
}
  
// Driver code
public static void Main(String []args)
{
    int n = 8;
  
    // Pre-compute the sequence
    precompute();
  
    Console.WriteLine(F[n]);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

47


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Improved By : princiraj1992, 29AjayKumar