How does the valid parentheses problem use a stack?

Iterate through the string character by character. When you encounter an opening bracket, push it onto the stack. When you encounter a closing bracket, check if the stack is empty, which means no matching opener, or if the top of the stack does not match the closing type. If either condition is true, the string is invalid. If it matches, pop the stack. After processing all characters, the string is valid only if the stack is empty, meaning every opener was matched. This runs in O of n time. The stack naturally handles nested brackets like parentheses inside square brackets inside curly braces.