Codeforces · 2183B

Yet Another MEX Problem

This C++ solution uses simulation for Codeforces 2183B Yet Another MEX Problem. Read the reasoning, inspect the code, or try your own test case below.

2183BArrays & prefix sumsSimulationC++27 lines
Solution240of 248
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Approach

Simulation

Yet Another MEX Problem matches the deletion process; the proven result min(array MEX, k-1) is what the code outputs.

Arrays & prefix sums

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

B_Yet_Another_MEX_Problem.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const long long INF = 1e17; //❄️
    #define int long long
    void solve() {
        int n, k; cin >> n >> k;
        vector<int> v(n);
        for (int i = 0; i < n; i++) cin >> v[i];
        vector<int> seen(n + 2); // mex is at most n + 1;
        for (int i = 0; i < v.size(); i++) {
            if (0 <= v[i] && v[i] <= n + 1) {
                seen[v[i]] = 1;
            }
        }
        int mex = 0;
        while (seen[mex]) mex++;
        cout << min(mex, k - 1) << '\n';
    }
    signed main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int t; cin >> t;
        while (t--) {
            solve();
        }
        return 0;
    }
        

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