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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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; ` `} `

Java

 `// 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 `

Python3

 `# 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 `

C#

 `// 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 `

Output:

```47
```

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Improved By : princiraj1992, 29AjayKumar