Baum Sweet Sequence

Baum Sweet Sequence is an infinite binary sequence of 0s and 1s. The nth term of the sequence is 1 if the number n has no odd number of contiguous zeroes in its binary representation, else the nth term is 0.

The first few terms of the sequence are:
b1 = 1 (binary of 1 is 1)
b2 = 0 (binary of 2 is 10)
b3 = 1 (binary of 3 is 11)
b4 = 1 (binary of 4 is 100)
b5 = 0 (binary of 5 is 101)
b6 = 0 (binary of 6 is 110)

Given a natural number n. The task is to find the nth term of the Baum Sweet sequence, i.e, check whether it contains any consecutive block of zeroes of odd length.

Input: n = 8
Output: 0
Explanations:
Binary representation of 8 is 1000. It
contains odd length block of consecutive 0s.
Therefore B8 is 0.

Input: n = 5
Output: 1

Input: n = 7
Output: 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to run a loop through the binary representation of n and count the length of all the consecutive zero blocks present. If there is at-least one odd length zero block, then the nth term for the given input n is 0 else it is 1.

 // CPP code to find the nth term of the // Baum Sweet Sequence #include using namespace std;    int nthBaumSweetSeq(int n) {     // bitset stores bitwise representation     bitset<32> bs(n);        // len stores the number of bits in the      // binary of n. builtin_clz() function gives      // number of zeroes present before the      // leading 1 in binary of n     int len = 32 - __builtin_clz(n);        int baum = 1; // nth term of baum sequence     for (int i = 0; i < len;) {         int j = i + 1;            // enter into a zero block         if (bs[i] == 0) {             int cnt = 1;                // loop to run through each zero block             // in binary representation of n             for (j = i + 1; j < len; j++) {                    // counts consecutive zeroes                  if (bs[j] == 0)                                        cnt++;                 else                     break;             }                // check if the number of consecutive             // zeroes is odd             if (cnt % 2 == 1)                 baum = 0;         }         i = j;     }        return baum; }    // Driver Code int main() {     int n = 8;     cout << nthBaumSweetSeq(n);     return 0; }

Output:

0

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