Consecutive Ones - Transition

At position $pos$:

$1.$ You can always place $0$. Recurse with prev = $0$.

$2.$ You can place $1$ only if prev $\neq 1$ and (not tight or $n[pos] = 1$). If prev was $1$ and you try to place $1$, you'd create $11$. Skip that branch. The tight logic works the same as base $10$: if tight and $n[pos] = 0$, you can't place $1$. This formula tells you how to compute dp(pos, tight, prev) from dp values at later positions. Getting this right is the core of DP.