Count number of binary strings without consecutive 1’s

Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1’s.

Examples:

Input:  N = 2
Output: 3
// The 3 strings are 00, 01, 10

Input: N = 3
Output: 5
// The 5 strings are 000, 001, 010, 100, 101

class Solution {
    // O(n) space solution
    public static int countStrings(int n) {
        // Let a[i] be the number of binary strings of length i which do not contain any two consecutive 1’s and which end in 0. 
        // Similarly, let b[i] be the number of such strings which end in 1.
        int a[] = new int[n];
        int b[] = new int[n];
        a[0] = b[0] = 1;
        for (int i = 1; i < n; i++) {
            a[i] = a[i - 1] + b[i - 1];
            b[i] = a[i - 1];
        }
        return a[n - 1] + b[n - 1];
    }

    // O(1) space solution
    public static int countStrings(int n) {
        if (n == 0)
            return 1;
        if (n == 1)
            return 2;
        int a0 = 1;
        int b0 = 1;
        for (int i = 1; i < n; i++) {
            int temp = a0 + b0;
            b0 = a0;
            a0 = temp;
        }
        return a0 + b0;
    }
}