M Me Mex#12332

M Me Mex (Editorial)

Problem Author:	Hadi Takla
Prepared by:	Hadi Takla
Edited by:	Hadi Takla

First , instead of choosing $n - 1$ n-1 arrays , we will take all $n$ arrays and then try to remove one.

Since we need to obtain Mex( $b$ ) = $k$ , $k$ should not be present in any chosen prefix.

Let $p_{i}$ p_i be the longest prefix from the array $a_{i}$ a_i that does not contain $k$ , and let $b = p_{1} + p_{2} + ... + p_{n}$ b = p_1 + p_2 + ... + p_n , and $f r e q_{x}$ freq_x is the frequency of integer $x$ in the array $b$ .

if we can't obtain a Mex equal to $k$ using all $n$ arrays answer is "NO".

Otherwise , we have all integers between $0$ and $k - 1$ k-1 present in the array $b$ .

Now , for each $i$ we will try to remove $p_{i}$ p_i from $b$ to see what happens.

for each integer $x \in p_{i}$ x \in p_i we will decrease $f r e q_{x}$ freq_x , if we got $f r e q_{x} = 0$ freq_x = 0 and $0 \leq x \leq k - 1$ 0 \le x \le k-1 then the array $i$ is necessary and we can't remove $p_{i}$ p_i.

after that increase $f r e q_{x}$ freq_x again and move to the next array.

if all $n$ arrays are necessary answer is "NO" ,otherwise answer is "YES".

Time complexity $O (m \cdot lo g (max (x)))$ \mathcal{O}(m \cdot \log(\max(x))) using map to track frequencies , $O (m)$ using frequency array.

$m = i = 1 \sum n sz_{i} = sz_{1} + sz_{2} + sz_{3} + ... + sz_{n}$ m=\displaystyle \sum_{i=1}^n \text{sz}_i=\text{sz}_1+\text{sz}_2+\text{sz}_3+...+\text{sz}_n.

#include<bits/stdc++.h>
#define endl '\n'
#define int long long
#define F first
#define S second
using namespace std;
const int N = 2e5 + 1;
const int M = 1e9 + 7;
signed main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int tc = 1;
    cin>>tc;
    while(tc--)
    {
        int n,k;
        cin>>n>>k;
        vector<int>p[n+1];
        for(int i=1;i<=n;i++)
        {
            int m;
            cin>>m;
            bool ok = 1;
            for(int j=1;j<=m;j++)
            {
                int x;
                cin>>x;
                if(x == k)
                ok = 0;
                if(ok)
                p[i].push_back(x);
            }
        }
        map<int,int>freq;
        for(int i=1;i<=n;i++)
        {
            for(auto it:p[i])
            freq[it]++;
        }
        int mex = 0;
        while(freq[mex])
        mex++;
        if(mex != k)
        {
            cout<<"NO"<<endl;
            continue;
        }
        int cnt = 0; //count of necessary arrays
        for(int i=1;i<=n;i++)
        {
            bool nec = 0;
            for(auto it:p[i])
            {
                freq[it]--;
                if(freq[it] == 0 && it < k)
                nec = 1;
            }
            cnt += nec;
            for(auto it:p[i])
            freq[it]++;
        }
        cout<<(cnt == n ? "NO" : "YES")<<endl;
    }
}