Leave the first valid top
Updated
After pop-while, either the stack is empty or peek() satisfies the condition. That remaining top is the nearest previous (or next, depending on scan direction) that passed. Do not pop it.
What to focus on
If you need to record an answer from that top, peek it. If you then push the current value, the new top is the current, and the valid one is buried again. That is intended.
Worked examples
kept
a = [1, 3, 5], after pop-while top>4- 3 is valid
- leave it
Result: peek 3
empty leftover
a = [5, 6], after pop-while top>1- no valid top
Result: []
then push
a = [1, 3], push 4- 4 is new top
- 3 still under
Result: [1, 3, 4]
Try it
Write leftoverTop(a, pred). a is the stack, written [bottom, ..., top]. pred is a function that takes one value and returns true or false. After pop-while, peek the leftover top if any. Then you may push the current value. After pop-while on [1, 3, 5] vs pred = top > 4, peek 3. After draining [5, 6], leftover is empty. Then push the current value.
The stack starts non-empty. Values are integers. pred is a function that takes one value and returns true or false. Empty leftover after pop-while is a legal outcome.