How does the next greater element problem use a monotonic stack?

Iterate through the array from right to left, maintaining a stack of elements in decreasing order from bottom to top. For each element, pop all elements from the stack that are smaller than or equal to the current element because they cannot be the next greater element for anyone to the left. The stack top after popping, if it exists, is the next greater element. Then push the current element. If the stack is empty after popping, there is no greater element to the right. Each element enters and leaves the stack at most once, giving O of n total time despite the nested loop appearance.