2226CUnreviewedunratedGPT-5.4 (high)

Mental Monumental (Easy Version)

Progressive hints first, then the full explanation and implementation when you're ready to cash out.

binary search data structures greedy math two pointers

Original problem

Review status

AI-generated and still unreviewed. Double-check the details before internalizing them.

Hints

Progressive nudges

Open only as much as you need to keep the solve alive.

First stop guessing and characterize a single number. For a fixed value $x$ , what values can $x \bmod b$ become as $b \ge 1$ varies? The whole problem is dead without this.

Suppose you want $x$ to become some smaller value $y < x$ . If $x \bmod b = y$ , then $x-y$ is a positive multiple of $b$ , and also $y < b$ . That already gives a very sharp inequality between $x$ and $y$ . Try proving: $y<x \text{ is reachable} \iff x>2y.$ Also, $x$ itself is always reachable by choosing $b>x$ .

Now fix a candidate answer $m$ . To have $\operatorname{mex}=m$ , you need to realize every value $0,1,\dots,m-1$ using distinct array elements. So this becomes a matching problem: an element $a_i$ can cover value $x$ iff $a_i=x \quad\text{or}\quad a_i\ge 2x+1.$

Here’s the exchange trick that makes the check sane instead of disgusting: if some value $x<m$ already exists in the array, you can always reserve one copy of $x$ to produce $x$ itself. Why? If some larger element was used to make $x$ , and a copy of $x$ was used to make some smaller $y$ , then swapping them still works because $x$ can make only very small values, while that larger element can make those too.

So for checking whether mex $m$ is possible:

Sort $a$ .
Remove (reserve) one copy of every present value in $[0,m-1]$ .
Let the missing values in $[0,m-1]$ be $x_1<x_2<\dots$ .
Let the remaining elements be sorted leftovers $v_1\le v_2\le\dots$ .
Greedily match each missing $x$ to the smallest leftover $v$ with $v\ge 2x+1.$ If every missing value gets matched, then mex $m$ is feasible. This predicate is monotone, so binary search the answer.

Observation 1: what can one element become?

Take one number $x$ . We choose any integer $b \ge 1$ , and the value becomes $x \bmod b$ .

What results are possible?

Getting $x$ itself is easy: choose any $b>x$ , then $x \bmod b = x$ .
For some smaller value $y<x$ , suppose $x \bmod b = y$ . Then $x-y = kb$ for some integer $k\ge 1$ . Also a remainder must satisfy $y<b$ . Therefore $x-y \ge b > y,$ so $x>2y.$

That gives a necessary condition.

Is it sufficient? Yep. If $x>2y$ , choose $b=x-y.$ Then $b>y$ , so the remainder is valid, and $x \bmod (x-y) = y.$

So the reachable values from $x$ are exactly $\{x\}\cup\{0,1,2,\dots,\lfloor (x-1)/2\rfloor\}.$

That’s the whole damn problem in one line.

Observation 2: checking whether mex $m$ is possible

To make $\operatorname{mex}=m$ , we need every value $0,1,\dots,m-1$ to appear after the operation.

Each array element can be used once, so this is a matching problem.

An element with value $v$ can produce a target value $x$ iff $v=x \quad\text{or}\quad v\ge 2x+1.$

So for a fixed $m$ , the question becomes:

Can we assign distinct array elements to all targets $0,1,\dots,m-1$ under that rule?

This feasibility is monotone: if we can make mex $m$ , then we can obviously make any smaller mex too. So binary search is fair game.

Observation 3: reserve exact copies first

This is the key exchange argument that saves us from writing cursed nonsense.

Suppose some value $x<m$ appears in the array. Then there exists an optimal construction where one copy of $x$ is left as $x$ .

Why is that always safe?

Assume in some solution a copy of $x$ is used to make some smaller value $y$ , and some other element $v$ is used to make $x$ .

Since $x$ is produced by $v$ and $v\ne x$ , we must have $v\ge 2x+1.$

Since the copy of $x$ makes $y<x$ , we know $x>2y.$ So $v\ge 2x+1 > 2y+1,$ meaning $v$ can also make $y$ .

Therefore we can swap roles:

use the copy of $x$ to make $x$ ,
use $v$ to make $y$ .

Nothing breaks.

So for a fixed $m$ :

for every present value $x\in[0,m-1]$ , reserve one copy of $x$ to cover $x$ itself;
the values in $[0,m-1]$ that are absent are the only ones we still need to create.

Now the remaining copies are just a pool of leftovers. A missing value $x$ can be made by a leftover value $v$ iff $v\ge 2x+1.$

Greedy check for a fixed $m$

After reserving one copy of each present value in $[0,m-1]$ :

let the missing targets be $x_1 < x_2 < \dots < x_k;$
let the leftover values, sorted, be $v_1 \le v_2 \le \dots \le v_t.$

We need to match each $x_i$ with a distinct leftover $v_j$ such that $v_j \ge 2x_i+1.$

The greedy strategy is standard and correct:

process missing values from small to large;
for each missing $x$ , take the smallest leftover value $v$ with $v\ge 2x+1$ .

Why? Because larger missing values only have stricter requirements. Burning a too-large leftover early is just self-sabotage. Classic greedy, classic not-being-an-idiot.

If at some step no leftover satisfies the threshold, then mex $m$ is impossible.

How to implement the check in $O(n)$ after sorting

Sort the array once.

For a given $m$ , scan the sorted array by groups of equal values. During this scan you can build two sorted lists directly:

missing: all values in $[0,m-1]$ that do not appear;
leftover: all array elements except one reserved copy of each present value in $[0,m-1]$ .

Both lists are naturally sorted. Then run a two-pointer greedy match.

So:

sorting: $O(n\log n)$ ,
one feasibility check: $O(n)$ ,
binary search over the answer: $O(\log n)$ checks.

Total per test case: $O(n\log n).$

Given $\sum n \le 2\cdot 10^5,$ this flies comfortably.

Correctness sketch

We proved:

A value $v$ can become exactly the set $\{v\}\cup [0,\lfloor (v-1)/2\rfloor].$
For any feasible mex $m$ , we may reserve one copy of every present $x<m$ to realize $x$ exactly.
Therefore the remaining task is only to cover missing values $x$ using leftovers with threshold $v\ge 2x+1.$
The smallest-suitable-leftover greedy is optimal for such threshold matching.

Hence the check is correct, and binary search over $m$ returns the maximum feasible mex, which is exactly $f(a)$ .

Complexity

For each test case:

sort once: $O(n\log n)$ ,
binary search on answer: $O(\log n)$ iterations,
each check: $O(n)$ .

So the total is $O(n\log n).$ Memory usage is $O(n).$

Nice, clean, no black magic, no overengineered garbage.

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int T;
    cin >> T;
    while (T--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (int &x : a) cin >> x;
        sort(a.begin(), a.end());

        auto can = [&](int m) -> bool {
            vector<int> missing;
            vector<int> leftover;
            missing.reserve(m);
            leftover.reserve(n);

            int need = 0; // next value in [0, m-1] we are looking at
            int i = 0;
            while (i < n) {
                int v = a[i];
                int j = i;
                while (j < n && a[j] == v) ++j;
                int cnt = j - i;

                while (need < m && need < v) {
                    missing.push_back(need);
                    ++need;
                }

                if (v < m && need == v) {
                    // reserve exactly one copy of v to realize v itself
                    ++need;
                    for (int k = 1; k < cnt; ++k) leftover.push_back(v);
                } else {
                    for (int k = 0; k < cnt; ++k) leftover.push_back(v);
                }

                i = j;
            }

            while (need < m) {
                missing.push_back(need);
                ++need;
            }

            int p = 0;
            int sz = (int)leftover.size();
            for (int x : missing) {
                int req = 2 * x + 1;
                while (p < sz && leftover[p] < req) ++p;
                if (p == sz) return false;
                ++p;
            }
            return true;
        };

        int lo = 0, hi = n + 1; // mex is at most n
        while (hi - lo > 1) {
            int mid = (lo + hi) / 2;
            if (can(mid)) lo = mid;
            else hi = mid;
        }

        cout << lo << '\n';
    }
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;

void setIO() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int main() {
    setIO();

    int T;
    cin >> T;
    while (T--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (int &x : a) cin >> x;
        sort(a.begin(), a.end());

        auto can = [&](int m) -> bool {
            vector<int> missing;
            vector<int> leftover;
            missing.reserve(m);
            leftover.reserve(n);

            int need = 0; // next value in [0, m-1] we are looking at
            int i = 0;
            while (i < n) {
                int v = a[i];
                int j = i;
                while (j < n && a[j] == v) ++j;
                int cnt = j - i;

                while (need < m && need < v) {
                    missing.push_back(need);
                    ++need;
                }

                if (v < m && need == v) {
                    // reserve exactly one copy of v to realize v itself
                    ++need;
                    for (int k = 1; k < cnt; ++k) leftover.push_back(v);
                } else {
                    for (int k = 0; k < cnt; ++k) leftover.push_back(v);
                }

                i = j;
            }

            while (need < m) {
                missing.push_back(need);
                ++need;
            }

            int p = 0;
            int sz = (int)leftover.size();
            for (int x : missing) {
                int req = 2 * x + 1;
                while (p < sz && leftover[p] < req) ++p;
                if (p == sz) return false;
                ++p;
            }
            return true;
        };

        int lo = 0, hi = n + 1; // mex is at most n
        while (hi - lo > 1) {
            int mid = (lo + hi) / 2;
            if (can(mid)) lo = mid;
            else hi = mid;
        }

        cout << lo << '\n';
    }
}

View on Codeforces Back to the list

Mental Monumental (Easy Version)

Progressive nudges

Full explanation

Observation 1: what can one element become?

Observation 2: checking whether mex $m$ is possible

Observation 3: reserve exact copies first

Greedy check for a fixed $m$

How to implement the check in $O(n)$ after sorting

Correctness sketch

Complexity

Generated C++

Mental Monumental (Easy Version)

Progressive nudges

Full explanation

Observation 1: what can one element become?

Observation 2: checking whether mex mmm is possible

Observation 3: reserve exact copies first

Greedy check for a fixed mmm

How to implement the check in O(n)O(n)O(n) after sorting

Correctness sketch

Complexity

Generated C++

Observation 2: checking whether mex $m$ is possible

Greedy check for a fixed $m$

How to implement the check in $O(n)$ after sorting