B is brute force
Check for every factor of n
calculate max min for each one and find the max difference between them

D is also easy, just count the frequency of each element
Then add fC2 for every frequency
and there is special case that 2^4 = 4^2 so just add the product of freq of 2 and freq of 4 😁❤️

E solution
Find the smallest element index and if the array is sorted from there then print its index otherwise if not sorted print -1

void code()
{
int n;
cin >> n;
vector<int> arr(n);
for (int &i : arr)
cin >> i;
int minIndex = 0;
for (int i = 0; i < n; i++)
if (arr[i] < arr[minIndex])
minIndex = i;
int res = minIndex;
for (int i = minIndex; i < n - 1; i++)
if (arr[i] > arr[i + 1])
res = -1;
cout << res << endl;
}

// E
// Enjoy ❤️

A and B done ✅ codeforces

C done

D done

E done 😍

F done

Dm fast, time's running 😍

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A done codeforces 😍

CF B done ✅

Leetcode D done 😍 dm fast

C done ✅ 😍

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Codeforces A done ✅