N th Fibonacci Number

Problem statement
The n-th term of Fibonacci series F(n), where F(n) is a function, is calculated using the following formula -

    F(n) = F(n - 1) + F(n - 2), 
    Where, F(1) = 1, F(2) = 1


Provided 'n' you have to find out the n-th Fibonacci Number. Handle edges cases like when 'n' = 1 or 'n' = 2 by using conditionals like if else and return what's expected.

"Indexing is start from 1"


Example :
Input: 6

Output: 8

Explanation: The number is ‘6’ so we have to find the “6th” Fibonacci number.
So by using the given formula of the Fibonacci series, we get the series:    
[ 1, 1, 2, 3, 5, 8, 13, 21]
So the “6th” element is “8” hence we get the output.
Detailed explanation ( Input/output format, Notes, Images )
Sample Input 1:
6


Sample Output 1:
8


Explanation of sample input 1 :
The number is ‘6’ so we have to find the “6th” Fibonacci number.
So by using the given formula of the Fibonacci series, we get the series:    
[ 1, 1, 2, 3, 5, 8, 13, 21]
So the “6th” element is “8” hence we get the output.


Expected time complexity :
The expected time complexity is O(n).


Constraints:
1 <= 'n' <= 10000     
Where ‘n’ represents the number for which we have to find its equivalent Fibonacci number.

Time Limit: 1 second




Solution :
//Java

import java.util.*;
public class Solution {
	public static void main(String[] args) {
		Scanner a=new Scanner(System.in);
		int n=a.nextInt();
		System.out.println(F(n));
	}
	int count=0;
	static int F(int n){
		int count=0;
		int ini=1;
		int tot=0;
		for(int i=1;i<=n;i++){
		if(n==1||n==2){
			count++;
		}
		else{
			tot=ini+count;
			ini=count;
			count=tot;
			}
		}
		return tot;
	}

}