Solution using frequency array:

Initialize an array of length $$$n$$$ with all the elements 0. This will be our frequency array. For every element we have in our list, we increase its frequency by one in the frequency array, $$$freq[arr[i]]+=1$$$. After we do this, we iterate through the frequency array and see how many elements have a frequency higher than 0 (it means they were mentioned in the list), let this number be $$$cnt$$$. Now if $$$cnt = n-1$$$ them we know for sure that who is the impostor, because all the other people are mentioned in the list. If $$$cnt ≠ n-1$$$ then we can't for sure know who is the impostor, so the answer is no.

Solution using set:

We introduce all the elements of the list in a set. A property of a set is that it doesn't contain a multiple elements of the same value. We just need to see how big the size of the set (how many distinct people are on the list). If this size is equal to n-1 then the answer is "YES", else "NO".

#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";
 	}
}

We iterate from the start and see how many carrots the first rabbit can eat, let this be $$$cnt_1$$$. We iterate from the end to see how many carrots the second rabbit can eat, let this be $$$cnt_2$$$. If $$$cnt_1 + cnt_2 \gt n$$$ then some carrots will be eaten by both rabbits, but this is not possible because they stop whenever they meet. The answer is $$$min(n, cnt_1+cnt_2)$$$.

#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];
	int Rabbit1 = 1;
 
	for(int i = 1 ;i < n;i++)
	{
		if(a[i] >= a[i-1]){
			Rabbit1++;
		}else{
			break;
		}
	}
 
	int Rabbit2 = 1;
 
	for(int i = n-2;i>=0;i--)
	{
		if(a[i] >= a[i+1]){
			Rabbit2++;
		}else{
			break;
		}
	}
	cout << min(n , Rabbit2 + Rabbit1);
}

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;
}

we can find the answer by looking at the cases depending on n % 4, or we can use a formula. In either case, below is proof of why it works.

#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";
	    }
	}
}

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

int main()
{
    long long n;
    cin >> n;
    long long cover = ((n / 2) * (n / 2) + 1) / 2;
    long long as = 4 * cover;
    long long bs = n * n - as;
    
    if (as == bs) cout << "Draw\n";
    else if (as < bs) cout << "Bob\n";
    else cout << "Alice\n";
}