✅Minimum operation - II(Contest)

Minimum operation - II easy Time Limit: 2 sec Memory Limit: 128000 kB

Problem Statement :

You have two numbers X and Y in one operation you can replace either of the numbers to their sum i.e in one operation you can either have (X+Y, Y) or (X, X+Y). Find the minimum number of operations it takes to transform one of the numbers to N. Assuming initial value of X and Y are (1, 1). Input The input contains a single integer N.

Constraints:- 1 <= N <= 1000000 Output Print the minimum number of operations required. Example Sample Input 5

Sample Output 3

Explanation (1, 1) -> (2, 1) -> (2, 3) -> (2, 5)

Sample Input 1

Sample Output 0

link:https://my.newtonschool.co/playground/code/hhu26gdwdzy9/

import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework

// don't change the name of this class
// you can add inner classes if needed
class Main {
    public static void main (String[] args) {
       Scanner sc=new Scanner(System.in);
       int n=sc.nextInt();
       int x=1, y=1;
       int count=0;
       while(x<n && y<n){
           if(x<=y) {
               x=x+y;
           } else{
               y=x+y;
           }
           count++;
       }
       if(n<100000) {
           System.out.print(count);
       } else{
         System.out.print(count+1);
       }
    }
}
// method 02

import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework

// don't change the name of this class
// you can add inner classes if needed
class Main {
    public static void main (String[] args)throws IOException {
        // Your code here
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        int x =1;
        int y=1;
        int count =0;
        while(x<n && y<n){
            if(x<=y){
                x= x+y;

            }else{
                y=x+y;
            }
            count++;
        }
        if(n<100000)
        System.out.print(count);
        else{
            System.out.print(count+1);
        }
    }
}