Related Articles

# Count non-adjacent subsets from numbers arranged in Circular fashion

• Last Updated : 07 Jul, 2021

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:
All possible subsets are {}, {1} and {2}.

Input: N = 3
Output: 4

Approach: Let’s find the answer to small values of N.
N = 1 -> All possible subsets are {}, {1}
N = 2 -> All possible subsets are {}, {1}, {2}
N = 3 -> All possible subsets are {}, {1}, {2}, {3}
N = 4 -> All possible subsets 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, 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 = 2;` `    ``// Starting two terms of the sequence``    ``F = 3;``    ``F = 4;` `    ``// Compute 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``;` `    ``// Compute 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``;` `    ``# Compute 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 = 2;` `    ``// Starting two terms of the sequence``    ``F = 3;``    ``F = 4;` `    ``// Compute 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`

## Javascript

 ``
Output:
`47`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up