I was solving 1839D - Ball Sorting which was a standard DP problem, when I submitted and got WA on test 1. This was so strange to me since I just pasted the input into my own IDE and the outputs were identical. But sure enough, the moment I submitted the same exact solution, the output was different. I had to spam multiple submissions onto Codeforces to see what went wrong.

WA soluion vs AC solution

280230092 280230005

The only difference between these two pieces of code were ++n; a[n]=n; and a[++n]=n;. At first I thought, shouldn't these have the exact same effect? That's what my compiler told me!

After another WA on test 1-to-debug submission, I realised that it is possible that Codeforces' compiler had put the original value of $$$n$$$ into some temporary memory first, and only then update $$$n$$$ and put the temporary value back into the array.

I have used this kind of assignment a[++n]=n so many times, I'm surprised that only now I'm discovering this! Which is so dangerous because imagine the devastation this could have caused me in offline contests where you only get one chance.

Is it bad practice to put any thing but a single variable into the index of an array? I hope to learn more from all of you.

Today I tried to solve 1932F - Feed Cats, and finally got Accepted, though not after some hard debugging. You can give a try at the problem first before reading this.

Here is my original submission with WA on test 3 268592520 and here is my AC submission 268597071.

My thought process went like this:

• I thought of a segment tree with sweepline style solution

• Though using segment tree with n=10^6 is very risky (high constant time)

• So I went ahead and compress the coordinates of the start point and end point of each segment.

• Query an addition at the compressed start point and a removal at the (compressed end point) +1.

This is WRONG, however. Since there is a chance of nothing happening at the compressed end point and something else happening at the (compressed end point) +1 that might influence the answer, this can give a wrong result. The really important part is the compressed (end point +1) where the removal ACTUALLY happens, which I failed to realise for 90 minutes.

Did I learn my lesson? Yes. Did you waste your time reading something you already know/ don't need to know yet? I hope not. Either way, perhaps I will leave this blog here as a cautionary tale for myself, to better tackle future problems I will have to solve.

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!