For a color $c$, define $S(c)$ as the first occurrence of $c$ and $E(c)$ as the last occurrence of $c$ in the input sequence.

Our first task is to recover the left and right bounds of each painted interval. For each color $c$, its interval must start at or before $S(c)$ and end on or after $E(c)$. In fact, we can assume the interval starts exactly at $S(c)$ and ends exactly on $E(c)$. To see why this is the case, note that if we had a procedure to recreate the final painting that involves painting a longer interval for $c$, we can substitute it with $[S(c), E(c)]$ without changing the end result.

Now, let’s investigate which configurations are valid. If $S(A) < S(B) < E(A) < E(B)$ for two colors A and B, we claim that the final configuration is impossible. Indeed, if A were painted first, then E(A) cannot have color A once B is painted. Similarly, we reach a contradiction assuming B is painted first.

Therefore, any two intervals $[S(A), E(A)]$ and $[S(B), E(B)]$ must be completely disjoint, or one must be contained within the other. It follows that the only colors that can appear inside an interval are the colors of the intervals it contains. All such configurations are valid paintings since we can always remove the shortest interval by painting it last. Suppose to the contrary that the shortest interval contained the start/end of another interval. Then it would have to contain that interval's end/start as well, and we have a contradiction!

Viewed another way, if we arrange all start and end points in order, replace start points by open braces (of the corresponding color) and end points by close braces (of the corresponding color), we must have a correct bracket sequence.

Finally, let's look at our painting process in reverse. In the last step, we paint (remove) all empty bracket sequences $[]$, and apply this procedure recursively. This shows that our final answer is the maximum depth of the bracket sequence.

Here is Mark Chen’s code:

#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

const int INF = 2000000000;

const int MAXN = 100005;
int color[MAXN], xmin[MAXN], xmax[MAXN], inStack[MAXN];

int n;

vector<int> stack;

int main() {
    scanf("%d", &n);

    fill(xmin, xmin + MAXN, INF);
    fill(xmax, xmax + MAXN, -INF);

    color[0] = color[n+1] = 0;

    for (int i = 0; i <= n+1; i++) {
        if (i >= 1 && i <= n) scanf("%d", &color[i]);
        xmin[color[i]] = min(xmin[color[i]], i);
        xmax[color[i]] = max(xmax[color[i]], i);

    int res = 0;  // max depth

    for (int i = 0; i <= n+1; i++) {
        int c = color[i];

        if (i == xmin[c]) {
            res = max(res, (int)stack.size());

        if (stack[stack.size()-1] != c) {
            cout << -1 << "\n";
            return 0;

        if (i == xmax[c]) {

    cout << res - 1 << "\n";
    return 0;