Valid Parenthesis String
Signals to notice
Brute force first
Try all 3^k possibilities for each *. Exponential in the number of wildcards. That direct path helps you understand the question, but it tends to treat every possibility as brand new instead of learning from earlier steps.
The key insight
Track a range [minOpen, maxOpen] of possible open-paren counts. '(' increases both. ')' decreases both. '*' decreases min, increases max. Clamp min to 0. Valid if min can reach 0 at end. Once you hold onto the right piece of information from moment to moment, the problem feels less like trial and error and more like following a shape that was there all along.
What must stay true
minOpen = minimum possible unmatched '(' (treat * as ')'). maxOpen = maximum possible unmatched '(' (treat * as '('). If maxOpen < 0, too many ')'. If minOpen == 0 at end, it's valid. When you keep that truth intact, each local choice supports the larger solution instead of fighting it.
Easy way to go wrong
Only tracking one count — you need a range because '*' creates branching. The range collapses the exponential branches into two linear bounds. The fix is usually to return to the meaning of each move, not just the steps themselves.