Use a minimum stack: each linked list node caches the minimum of all values from that node down to the bottom. This handles duplicate minima without searching when one is removed.
class Stack {
  private class Node(val value: Int, val least: Int, val next: Node)
  private var top: Node = null

  def push(x: Int): Unit = {
    val m = if (top == null) x else math.min(x, top.least)
    top = new Node(x, m, top)
  }

  def pop(): Int = {
    require(!isEmpty)
    val x = top.value
    top = top.next
    x
  }

  def isEmpty: Boolean = top == null

  def min: Int = {
    require(!isEmpty)
    top.least
  }
}
The representation invariant is that the chain from top is finite and acyclic, and every reachable node q satisfies
(1)
The abstraction function takes the values along this chain in top-to-bottom order. Thus pushing changes to , and popping a nonempty returns its head and leaves its tail. The cached-minimum equation follows for a new node from the previously valid chain; removing a head leaves every surviving equation unchanged. Consequently
(2)
The bounds use the usual constant-cost allocation and integer arithmetic model. Storage is linear in the number of entries. min and pop require a nonempty stack; min leaves it unchanged.

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