Given an array A,compute B s.t B[i] stores the nearest element to the left of A[i] which is smaller than A[i]

Question

Given an array A[1..n], we want to compute another array B[1..n] such that B[i] stores the nearest element to the left of A[i]

Answer 1

B[1]=A[1]
push(B[1])
for i=2 to n do
{
    while(A[i] > stack_top ANS stack_top!=NULL)
       pop()
    if(stack_top=NULL)
        B[i]=A[i]
    else
        B[i]=stack_top
    push(A[i])
}

As IVlad pointed out that each element is pushed and poped atmost once, time is O(n).

pl do correct me if there is some mistake, and I am curious for any alternate solution which avoids stack and is cheaper.