Monotone stack application

Previous larger element

For every input element a[i], the algorithm finds its rightmost preceding element with a strictly larger value.

Positive integers separated by spaces or commas
Big step one item
Shift + ← →
Small step one operation
← →

Stack

READY

in stack current being removed

Pseudocode

for i = 1 … n
  while S not empty and S.top ≤ a[i]
    S.pop()
  b[i] = S.empty ? ∅ : S.top
  S.push(i)