Weird behavior (order of execution)

Problems and Contests
#1

I redid USACO Silver Paired up. My approach is to pair the cow with smallest output with the cow with largest output as long as possible. I implement this using standard two pointers. The key loop which gives an error follows.

while (i <= j) {
    ans = max(ans, pairs[i].f + pairs[j].f);
    int amt = min(pairs[i].s, pairs[j].s);
    pairs[i].s -= amt;
    pairs[j].s -= amt;
    if (pairs[i].s == 0) ++i;
    if (pairs[j].s == 0) --j;
    //cout << pairs[i].s << " " << pairs[j].s << endl;
  }

In the sample input, we have i = j = 1 and pairs[i].s = pairs[j].s = 2. Then amt = 2 and the pairs[i].s and pairs[j].s should both come out to be 2-2 = 0.

If I do the following 4 lines in this order:

pairs[i].s -= amt;
pairs[j].s -= amt;
if (pairs[i].s == 0) ++i;
if (pairs[j].s == 0) --j;

their values pairs[i].s and pairs[j].s are instead -2. The program results in an infinite loop.

However, if I change the order of the lines:

pairs[i].s -= amt;
if (pairs[i].s == 0) ++i;
pairs[j].s -= amt;
if (pairs[j].s == 0) --j;

the program gives the correct answer.

Why did simply separating the two subtraction lines (pairs[i].s -= amt and pairs[j].s -= amt) make a difference? The first order of execution seems to cause obstructive implicit behavior that I don’t follow.

#2

Which of the two orders do you find confusing?

As I mentioned here, subtracting 2 from 2 twice gives -2, so it should be clear why the first order behaves as it does.

#3

I understand now. I kept thinking the same behavior that applies when i \ne j would apply when they are equal. When i = j, we want to pair half the cows with the other half, so I need to divide amt by 2. Since I did not do this, I subtracted 2 from the same value twice.

I unintentionally cured this in the latter order because i would increase since pairs[i].s = 0 after removing 2 once. However, I should avoid these ingenuine corrections and stick to the legitimate way of handling the error. In this case, I could have delved deeper into the i = j case by replacing j with i in my code. Then I could realize that we’re subtracting amt from pairs[i].s twice when we only need to subtract amt/2 each time.

Thank you!