Count Pairs Of Consecutive Zeros - GeeksforGeeks

Count Pairs Of Consecutive Zeros - GeeksforGeeks
Consider a sequence that starts with a 1 on a machine. At each successive step, the machine simultaneously transforms each digit 0 into the sequence 10 and each digit 1 into the sequence 01.
After the first time step, the sequence 01 is obtained; after the second, the sequence 1001, after the third, the sequence 01101001 and so on.
How many pairs of consecutive zeros will appear in the sequence after n steps?

Input : Number of steps = 3
Output: 1
// After 3rd step sequence will be  01101001

Input : Number of steps = 4
Output: 3
// After 4rd step sequence will be 1001011001101001

Input : Number of steps = 5
Output: 5
// After 3rd step sequence will be  01101001100101101001011001101001

This is a simple reasoning problem. If we see the sequence very carefully , then we will be able to find a pattern for given sequence. If n=1 sequence will be {01} so number of pairs of consecutive zeros are 0, If n = 2 sequence will be {1001} so number of pairs of consecutive zeros are 1, If n=3 sequence will be {01101001} so number of pairs of consecutive zeros are 1,
If n=4 sequence will be {1001011001101001} so number of pairs of consecutive zeros are 3.

So length of the sequence will always be a power of 2. We can see after length 12 sequence is repeating and in lengths of 12. And in a segment of length 12, there are total 2 pairs of consecutive zeros. Hence we can generalize the given pattern q = (2^n/12) and total pairs of consecutive zeros will be 2*q+1.

int consecutiveZeroPairs(int n)

{

    // Base cases

    if (n==1)

        return 0;

    if (n==2 || n==3)

        return 1;

 

    // Calculating how many times divisible by 12, i.e.,

    // count total number repeating segments of length 12

    int q = (pow(2,n) / 12);

 

    // number of consecutive Zero Pairs

    return 2*q + 1;

}

Read full article from Count Pairs Of Consecutive Zeros - GeeksforGeeks