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Thursday, 24 January 2013

Fibonacci numbers - 4 solutions

Find the n-th Fibonacci number. The Fibonacci numbers are integers 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
Wikipedia says:
By definition, the first two numbers in the Fibonacci sequence are 0 and 1 (alternatively, 1 and 1), and each subsequent number is the sum of the previous two. 
For example the 7-th Fibonacci number in the sequence is 8. Here are a few solutions. The first one is a trivial one and follows the definition - i.e. it uses recursion.

long fibonacciRecursive(int n) {
    if (n == 0 || n == 1) {
        return n;
    }
    return fibonacciRecursive(n - 1) + fibonacciRecursive(n - 2);
}


The second solution is iterative. A drawback here is that the size of the array which is used to store the Fibonacci numbers needs to be defined statically.

long fibonacciIterative(int n) {
    int size = 100;
    long[] fibonacciNumbers = new long[size + 1];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;
  
    for (int i = 2; i <= n; i++) {
        fibonacciNumbers[i] = fibonacciNumbers[i - 1]
                            + fibonacciNumbers[i - 2];
    }
    return fibonacciNumbers[n];
}


The third solution is again iterative. The difference here is that it uses an array with size 3. This is a very elegant and optimized solution.

long fibonacciIterativeLowMem(int n) {
    int[] fibonacciNumbers = new int[3];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;

    for (int i = 2; i <= n; i++) {
        fibonacciNumbers[i % 3] = fibonacciNumbers[(i - 1) % 3]
                                + fibonacciNumbers[(i - 2) % 3];
    }
    return fibonacciNumbers[n % 3];
}


Here is one more solution that uses an optimization called memoization. The idea here is that we avoid the calculation of the tasks we already had calculated.
long fibonacciMemoization(int n) {
    int size = 100;
    long[] fibonacciNumbers = new long[size+1];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;
    for (int i = 2; i < size; i++) {
        fibonacciNumbers[i] = -1;
    }
  
    if (fibonacciNumbers[n] != -1) {
        return fibonacciNumbers[n];
    } else {
        fibonacciNumbers[n] = fibonacciMemoization(n - 1)
                            + fibonacciMemoization(n - 2);
        return fibonacciNumbers[n];
    }
}

2 comments:

  1. One more,

    Using matrix multiplication. You can find nth fibonacci number in O(log n) time.

    ReplyDelete
    Replies
    1. Thanks, Dhaval!
      http://bit.ly/1bO4mjF

      Delete