# Number of ways an array can be filled with 0s and 1s such that no consecutive elements are 1

Given a number N, find the number of ways to construct an array of size N such that it contains only 1s and 0s but no two consecutive indexes have value 1 in them.

Examples:

Input : 2 Output : 3Explanation:For n=2, the possible arrays are: {0, 1} {1, 0} {0, 0} Input : 3 Output : 5Explanation:For n=3, the possible arrays are: {0, 0, 0} {1, 0, 0} {0, 1, 0} {0, 0, 1} {1, 0, 1}

**Naive Approach:**

The basic brute force approach would be to construct all the possible ways that the array can be filled with 1s and 0s, and then checking if there are any two consecutive 1s in the array if there are, do not count those arrays.

Since each element has 2 possible values, 1 and 0, and there are n total elements, the total number of arrays without any restriction will be of exponential order i.e 2^{n}.

**Efficient Approach:**

If we observe a bit closely, we can notice that there is a pattern forming in the input and output.

For **n = 1**, number of ways is **2** i.e. {0}, {1}

for **n = 2**, number of ways is **3**

Similarly,

for **n = 3** number of ways is **5**

for **n = 4** number of ways is **8**

and so on…

Let T() be the function which gives the number of ways the array of size n can be filled, then we get the following recurrence relation

T(n) = T(n-1) + T(n-2)

And this is the recurrence relation of **Fibonacci series<**.

Hence, the output for any n is equal to the **(n+2) ^{th} term** of the Fibonacci series starting from 1.

i.e. 1 1 2 3 5 8 11…

So now we just need to compute the Fibonacci sequence up to the (n+2)^{th} elements and that will be the answer.

**Time complexity is O(n)**

## C++

`// C++ implementation of the ` `// above approach ` `#include <iostream> ` `using` `namespace` `std; ` ` ` ` ` `// The total number of ways ` ` ` `// is equal to the (n+2)th ` ` ` `// Fibonacci term, hence we ` ` ` `// only need to find that. ` ` ` `int` `nth_term(` `int` `n) ` ` ` `{ ` ` ` `int` `a = 1, b = 1, c = 1; ` ` ` ` ` `// Construct fibonacci upto ` ` ` `// (n+2)th term the first ` ` ` `// two terms being 1 and 1 ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `c = a + b; ` ` ` `a = b; ` ` ` `b = c; ` ` ` `} ` ` ` ` ` `return` `c; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `int` `main() ` ` ` `{ ` ` ` ` ` `// Take input n ` ` ` `int` `n = 10; ` ` ` `int` `c = nth_term(n); ` ` ` ` ` `// printing output ` ` ` `cout << c; ` ` ` `} ` ` ` `// This code is contributed by Sumit Sudhakar. ` |

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## Java

`// Java implementation of the ` `// above approach ` `class` `Main ` `{ ` ` ` ` ` `// The total number of ways ` ` ` `// is equal to the (n+2)th ` ` ` `// Fibonacci term, hence we ` ` ` `// only need to find that. ` ` ` `public` `static` `int` `nth_term(` `int` `n) ` ` ` `{ ` ` ` `int` `a = ` `1` `, b = ` `1` `, c = ` `1` `; ` ` ` ` ` `// Construct fibonacci upto ` ` ` `// (n+2)th term the first ` ` ` `// two terms being 1 and 1 ` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{ ` ` ` `c = a + b; ` ` ` `a = b; ` ` ` `b = c; ` ` ` `} ` ` ` ` ` `return` `c; ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `// Take input n ` ` ` `int` `n = ` `10` `; ` ` ` `int` `c = nth_term(n); ` ` ` ` ` `// printing output ` ` ` `System.out.println(c); ` ` ` `} ` `} ` |

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## Python3

`# Python3 implementation of ` `# the above approach ` ` ` `# The total number of ways ` `# is equal to the (n+2)th ` `# Fibonacci term, hence we ` `# only need to find that. ` `def` `nth_term(n) : ` ` ` ` ` `a ` `=` `1` ` ` `b ` `=` `1` ` ` `c ` `=` `1` ` ` ` ` `# Construct fibonacci upto ` ` ` `# (n+2)th term the first ` ` ` `# two terms being 1 and 1 ` ` ` `for` `i ` `in` `range` `(` `0` `, n) : ` ` ` `c ` `=` `a ` `+` `b ` ` ` `a ` `=` `b ` ` ` `b ` `=` `c ` ` ` `return` `c ` ` ` `# Driver Code ` ` ` `# Take input n ` `n ` `=` `10` `c ` `=` `nth_term(n) ` ` ` `# printing output ` `print` `(c) ` `# This code is contributed by ` `# Manish Shaw (manishshaw1) ` |

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## C#

`// C# implementation of the ` `// above approach ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// The total number of ways ` ` ` `// is equal to the (n+2)th ` ` ` `// Fibonacci term, hence we ` ` ` `// only need to find that. ` ` ` `static` `int` `nth_term(` `int` `n) ` ` ` `{ ` ` ` `int` `a = 1, b = 1, c = 1; ` ` ` ` ` `// Construct fibonacci upto ` ` ` `// (n+2)th term the first ` ` ` `// two terms being 1 and 1 ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{ ` ` ` `c = a + b; ` ` ` `a = b; ` ` ` `b = c; ` ` ` `} ` ` ` ` ` `return` `c; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` ` ` `// Take input n ` ` ` `int` `n = 10; ` ` ` `int` `c = nth_term(n); ` ` ` ` ` `// printing output ` ` ` `Console.WriteLine(c); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007 ` ` ` ` ` ` ` |

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## PHP

`<?php ` `// PHP implementation of the ` `// above approach ` ` ` ` ` `// The total number of ways ` ` ` `// is equal to the (n+2)th ` ` ` `// Fibonacci term, hence we ` ` ` `// only need to find that. ` ` ` `function` `nth_term(` `$n` `) ` ` ` `{ ` ` ` `$a` `= 1; ` `$b` `= 1; ` `$c` `= 1; ` ` ` ` ` `// Construct fibonacci upto ` ` ` `// (n+2)th term the first ` ` ` `// two terms being 1 and 1 ` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++) ` ` ` `{ ` ` ` `$c` `= ` `$a` `+ ` `$b` `; ` ` ` `$a` `= ` `$b` `; ` ` ` `$b` `= ` `$c` `; ` ` ` `} ` ` ` ` ` `return` `$c` `; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` ` ` `// Take input n ` ` ` `$n` `= 10; ` ` ` `$c` `= nth_term(` `$n` `); ` ` ` ` ` `// printing output ` ` ` `echo` `$c` `; ` ` ` `// This code is contributed by nitin mittal ` `?> ` |

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**Output:**

144

We can further optimize above solution to work in O(Log n) using matrix exponentiation solution for finding n-th Fibonacci number (Please see methods 4, 5 and 6 of Program for Fibonacci numbers ).

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