1 9 7 5 4 1 7 5 4 9 1 5 4 7 9 1 4 5 7 9

The 9 moved to the right very quickly, but the 4 moved to the left very slowly.

In general, a single element can move left at most once per iteration of bubble sort.

Let's "normalize" the input array to be a permutation of 0 ... n-1, so that the value of each array element indices the index where that element should go in the sorted array:

1 9 7 5 4 | v 0 4 3 2 1

For any $i$, the value $i - a_i$ is therefore a lower
bound on the total number of bubble passes that need to occur before the array
is sorted. (If positive, it means the element needs to move left by that amount.
If negative, it's *definitely* a lower bound.)

Now, can we say that $\max_i(i - a_i)$ is actually the answer? Indeed, we can! However, to prove this requires some subtlety, since of course some elements can move right, thus some elements have increasing values of $i - a_i$.

To see why this is actually true, consider any element $a_i$ which has a positive value of $i - a_i$ (i.e., that element is to the right of where it should be). Then there must be some element to the left which is greater; then we can check from the way bubble sort operates that our element $a_i$ must move left. So $i - a_i$ decreases by 1.

Next, consider any element $a_i$ where $i - a_i = 0$ (that is, the element is in exactly the correct position.) It may or may not move left, but it certainly will not move right: for it to move to the right, the element just right of it must be smaller, but then some element to its left must be greater than it, and in that case the element will move left instead.

Therefore, $\max_i(i - a_i)$, if positive, will decrease by 1 in a bubble sort iteration.

Note that after counting the number of bubbles needed to reach the sorted array, we need to add 1 to account for the final iteration of the algorithm in the given pseudocode.

#include <cstdio> #include <algorithm> using namespace std; struct Entry { int index; int value; }; Entry entries[100000]; int main() { int n; scanf("%d", &n); for (int i = 0; i < n; i++) { entries[i].index = i; scanf("%d", &entries[i].value); } sort(entries, entries + n, [](Entry a, Entry b) { // Break ties by making the smaller element be whichever // element was first in the array originally. return a.value < b.value || (a.value == b.value && a.index < b.index); }); int answer = 0; for (int j = 0; j < n; j++) { // In terms of the notation from the above analysis, we have, // entries[j].index = i // j = a_i answer = max(answer, entries[j].index - j); } printf("%d\n", answer + 1); }