[Coding Made Simple] Number without consecutive 1s in binary representation

Given a number n, find the total number of numbers from 0 to 2^n - 1 which do not have consecutive 1s in their binary representation.

 

Solution.  This problem is the equivalence of fibonacci sequence. For a given n, f(n) = f(n - 1) + f(n - 2).

Analysis:  Denote f(n) as the answer of number n. 

to get numbers of n bits from n - 1 bits,  we add a prefix 0 or 1 to all n - 1 bits numbers.  Adding a prefix 0 has no impact on consecutive 0s, so the total number of non-consecutive 1s in all n bits with a leading 0 prefix is just f(n - 1); Adding a prefix 1 forces the 2nd MSB be 0. With the leading two bits fixed as 10, the total number of non-consecutive 1s in all n bits with a leading 1 prefix is just f(n - 2).

 

public int getTotalNumberOfNoConsecutiveOnes(int n) {
    if(n == 0) {
        return 1;
    }
    if(n == 1) {
        return 2;
    }
    int[] T = new int[n + 1];
    T[0] = 1;
    T[1] = 2;
    for(int i = 2; i <= n; i++) {
        T[i] = T[i - 1] + T[i - 2];
    }
    return T[n];
}