why do you suppose in the case of one that the next number should be always one?
so that the we make the next call by passing (n-2)
it might be possible that the next number to be 0?!
Optimal binary string recursion
hello @mzk1994
yeah thats why we broke it into two cases.
a) we place 0 at nth place so simply we can call f(n-1)
b) we place 1 at nth place,then clearly we cannot place 1 again at n-1 th place becuase two consecutive 1’s not allowed. that means at n-1th place we will always have 0.
and becuase values at n and n-1 position are already fixed, we will call on remaining f(N-2)
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