The problem is just implementation. Check if exactly $$$n-1$$$ numbers are encountered. You can easily do this using a set or a frequency array.

#include "bits/stdc++.h"
using namespace std;
int main()
{
    set<int> s;
 	int n, m;
 	cin >> n >> m;
 	for(int i = 0;i < m;i++){
 		int x;
 		cin >> x;
 		s.insert(x);
 	}
 	if(s.size() == n - 1){
 		cout << "YES";
 	}else{
 		cout << "NO";
 	}
}

The array we make should consist only of the median of the array (the middle element after sorting the array). If $$$n$$$ is even and there are two medians, both medians and all values between them are optimal choices.

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

int main()
{
    int n;
    cin >> n;
    int a[n];
    for(int i = 0;i < n;i++){
    	cin >> a[i];
    }
    sort(a,a+n);
    long long ans = 0;
    int median = a[n/2];
    for(int i = 0;i < n;i++){
    	ans += abs(median - a[i]);
    }
    cout << ans;
}

We need to find if the difference between $$$n$$$ and $$$x$$$ is also a permutation of $$$x$$$'s digits. There are many ways to do this, and our solution is to transform $$$x$$$ and $$$n-x$$$ into strings, then sort the strings and see if they are equal. This is possible in C++ using to_string(int value) function.

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

int main()
{
    long long n,m;
    cin >> n >> m;
    m-=n;
    if(m <= 0){
        cout << "NO";
        return 0;
    }
    string s = to_string(n);
    string k = to_string(m);
    sort(s.begin(), s.end());
    sort(k.begin(), k.end());
    if(k==s){
        cout << "YES\n";
    }else{
        cout << "NO\n";
    }
}

We iterate through string $$$s$$$, and whenever we encounter a substring $$$t$$$, we add to the answer the number of unique substrings we can make using it. We can calculate this with the formula (number of characters since the start of the last encounter)$$$\cdot$$$(the number of characters until the end of the string). This is because the start of the substring must be between the start of the last $$$t$$$ substring in $$$s$$$ and the start of the $$$t$$$ substring just encountered(so we don't double count previous substrings). The end must be between the last element of this $$$t$$$ substring and the last element of string $$$s$$$.

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

int main()
{
    string s, t;
    long long n;
    cin >> n >> s >> t;
    long long l = 0;
    long long ans = 0;
    for(int i = 0; i < n-1; i++)
    {
        if(s[i] == t[0] && s[i+1] == t[1])
        {
            long long from = i-l+1;
            long long to = n-(i+1);
            ans += from*to;
            l = i + 1;
        }
    }
    cout << ans << endl;
}

If $$$n$$$ % $$$4$$$ $$$=$$$ $$$2$$$ Alice wins, if not and $$$n$$$ % $$$2$$$ $$$=$$$ $$$0$$$ it is a draw and otherwise Bob wins.

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

int main()
{
	int t;
	cin >> t;
	while(t--)
	{
	    int n;
	    cin >> n;
	    if(n%4==2){
	    	cout << "Alice\n";
	    }else if(n%2==0){
	    	cout << "Draw\n";
	    }else{
	    	cout << "Bob\n";
	    }
	}
}