### hxano's blog

By hxano, history, 6 weeks ago,

The problem goes as follows:

You are given an array a of N non-negative integers, indexed from 1 to N. Define prev(x) such that prev(x) is the largest integer that satifies both following conditions: prev(x) < x, and a[prev(x)] < x. Print prev(i) for each i from 1 to N.

This is a basic problem that can be solved using stacks. However, I've seen an interesting implementation that avoids data structures altogether. Here's the main code for it:

cin>>n;
for (int i=1; i<=n; ++i) cin>>a[i];
a[0]=-inf;
for (int i=1; i<=n; ++i){
int p=i-1;
while (a[p]>=a[i]) p=sol[p];
sol[i]=p;
}
for (int i=1; i<=n; ++i) cout<<sol[i]<<" "; cout<<"\n";
return 0;


It seems to me that this runs in O(n) on average, but I cannot find its worst-case complexity. I suspect it to be O(n^2) but have had no luck proving it so far (I once used this implementation in a practice problem and got TLEed). Any help would be greatly appreciated. Thanks!

• +8

 » 6 weeks ago, # |   +18 It is doing the same thing as the stack solution, using the sol array to keep the stack
•  » » 6 weeks ago, # ^ |   0 Do you have a proof for it?
•  » » » 6 weeks ago, # ^ |   +13 Proof of the complexity would be the same as for the stack. You do n pushes into the stack and each element is removed at most once, so the number of removals is at most n, thus the complexity is O(n).
 » 6 weeks ago, # |   +13 not a proof, but if you clear the stack at the end, then number of iterations is exactly n  int iterations = 0; for (int i=1; i<=n; ++i){ int p=i-1; while (a[p]>=a[i]) p=sol[p], iterations++; sol[i]=p; } //clearing the stack at the end int p = n; while (p >= 1) p=sol[p], iterations++; assert(iterations == n);