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. In this article, we will learn how to print Fibonacci Series in Java up to the N term where N is the given number.
Examples of Fibonacci Series in Java
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.
Methods to Print Fibonacci Series in Java
There are many methods to print Fibonacci Series in Java as mentioned below:
- Iterative Approach
- Recursive Approach
- Using Memoization
- Using D.P.(Dynamic Programming)
1. Iterative Approach to Print Fibonacci Series in Java
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 numbers 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 approachclass 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
The complexity of the above method
Time Complexity: O(N)
Auxiliary Space: O(1)
2. Fibonacci Series Using Recursion in Java
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 the previous two values 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 values as:
recursive_function(N – 1) + recursive_function(N – 2);
Below is the implementation of the above approach:
Java
// Recursive implementation of// Fibonacci Seriesclass 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
The complexity of the above method
Time Complexity: O(2N)
Auxiliary Space: O(1)
3. Fibonacci Series Using Memoization
In the above example, Its time complexity is O(2n) which can reduce to O(n) using the memoization technique which will help to optimize the recursion method. This is because the function computes each Fibonacci number only once and stores it in the array.
Below is the implementation of the above approach:
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static void main(String[] args) { int n = 10; // Number of terms to print int[] memo = new int[n + 1]; for (int i = 1; i <= n; i++) { System.out.print(fibonacci(i, memo) + " "); } } public static int fibonacci(int n, int[] memo) { if (memo[n] != 0) return memo[n]; if (n == 1 || n == 2) return 1; else { memo[n] = fibonacci(n - 1, memo) + fibonacci(n - 2, memo); return memo[n]; } }} |
1 1 2 3 5 8 13 21 34 55
The Complexity of the above method
Time Complexity: O(n)
Auxiliary Space: O(n)
4. Printing Fibonacci Series Using Dynamic Programming
We can avoid the repeated work done in method 2 by storing the Fibonacci numbers calculated so far.
The steps to writing the Program are mentioned below:
- 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 Seriesclass 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
The complexity of the above method
Time Complexity: O(N)
Auxiliary Space: O(N)
To know more about the Fibonacci Series, refer to the article – Program to print first n Fibonacci Numbers.
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