Given a number N, we need to find the Fibonacci Series up to the N term.
The Fibonacci series is a series of elements where, the previous two elements are added to get the next element, starting with 0 and 1.
Examples:
Input: N = 10
Output: 0 1 1 2 3 5 8 13 21 34
Here first term of Fibonacci is 0 and second is 1, so that 3rd term = first(o) + second(1) etc and so on.Input: N = 15
Output: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377
Method 1 – Iterative: Initialize the first and second numbers to 0 and 1. Following this, we print the first and second numbers. Then we send the flow to the iterative while loop where we get the next number by adding the previous two number and simultaneously we swap the first number with the second and the second with the third.
Below is the implementation of the above approach:
Java
// Java program for the above approach class GFG { // Function to print N Fibonacci Number static void Fibonacci(int N) { int num1 = 0, num2 = 1; int counter = 0; // Iterate till counter is N while (counter < N) { // Print the number System.out.print(num1 + " "); // Swap int num3 = num2 + num1; num1 = num2; num2 = num3; counter = counter + 1; } } // Driver Code public static void main(String args[]) { // Given Number N int N = 10; // Function Call Fibonacci(N); } } |
0 1 1 2 3 5 8 13 21 34
Time Complexity: O(N)
Auxiliary Space: O(1)
Method 2 – Using Recursion: Since Fibonacci Number is the summation of the two previous numbers. We can use recursion as per the following condition:
- Get the number whose Fibonacci series needs to be calculated.
- Recursively iterate from value N to 1:
- Base case: If the value called recursively is less than 1, the return 1 the function.
-
Recursive call: If the base case is not met, then recursively call for previous two value as:
recursive_function(N – 1) + recursive_function(N – 2);
- Return statement: At each recursive call(except the base case), return the recursive function for the previous two value as:
recursive_function(N – 1) + recursive_function(N – 2);
Below is the implementation of the above approach:
Java
// Recursive implementation of // Fibonacci Series class GFG { // Function to print the fibonacci series static int fib(int n) { // Base Case if (n <= 1) return n; // Recursive call return fib(n - 1) + fib(n - 2); } // Driver Code public static void main(String args[]) { // Given Number N int N = 10; // Print the first N numbers for (int i = 0; i < N; i++) { System.out.print(fib(i) + " "); } } } |
0 1 1 2 3 5 8 13 21 34
Time Complexity: O(2N)
Auxiliary Space: O(1)
Method 3 – Using Dynamic Programming: We can avoid the repeated work done in method 2 by storing the Fibonacci numbers calculated so far. Below are the steps:
- Create an array arr[] of size N.
- Initialize arr[0] = 0, arr[1] = 1.
- Iterate over [2, N] and update the array arr[] as:
arr[i] = arr[i – 2] + arr[i – 1]
- Print the value of arr[N].
Below is the implementation of the above approach:
Java
// Dynamic Programming approach for // Fibonacci Series class fibonacci { // Function to find the fibonacci Series static int fib(int n) { // Declare an array to store // Fibonacci numbers. // 1 extra to handle case, n = 0 int f[] = new int[n + 2]; int i; // 0th and 1st number of // the series are 0 and 1 f[0] = 0; f[1] = 1; for (i = 2; i <= n; i++) { // Add the previous 2 numbers // in the series and store it f[i] = f[i - 1] + f[i - 2]; } // Nth Fibonacci Number return f[n]; } public static void main(String args[]) { // Given Number N int N = 10; // Print first 10 term for (int i = 0; i < N; i++) System.out.print(fib(i) + " "); } } |
0 1 1 2 3 5 8 13 21 34
Time Complexity: O(N)
Auxiliary Space: O(N)
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