if you have such a approach it would be great if you could share

Yeah i tried greedy but it didn't work

Just assume it is $$$1 \le n,m \le 10^6$$$. There are bunch of easy $$$O(nlogn)$$$ solutions like this problem. It is almost the same, asking to divide array into subarrays with limitation on the size but cost of subarray is only max insted of max-min.

I also know one meme problem, with $$$n \le 2.5 * 10^6$$$ and $$$tl =200ms$$$ to force the $$$O(n)$$$ solution.

I think the $$$O(nlogn)$$$ can be a good challenging problem for people in range of mid~high expert (for reference I solved the first problem in 2.5 hours inside an onsite contest when my rating was swinging between 1890 ~ 1949, my friend who was IM on cf at that point of time solved it in 20mins), but $$$O(n)$$$ is much harder and require good ds skills.

don't mind those boring people who kept downvoting.

I think if a person downvote some topics which are not meaningless, he is noob

Oh, I guess I got it. It's meaningless to have order. So I think it's always better to take combinations in non-decreasing order.

I'm not sure the code below is right or not. you can test it with your input like:

5 2

5 11 13 4 12

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

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<int> vec(n);
    for (int &num : vec)
    {
        cin >> num;
    }
    std::sort(vec.begin(), vec.end());
    std::vector<int> dp(n+1, 0);
    for (int i = m; i <= n; i++)
    {
        dp[i] = vec[i - 1] - vec[0];
        for (int j = m; j <= i - m; j++)
        {
            dp[i] = std::min(std::max(dp[j], vec[i - 1] - vec[j]), dp[i]);
        }
    }
    cout << dp[n] << endl;
}

int main()
{
    // freopen("input.txt", "r", stdin);
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}

THANK YOU SO MUCH for the suggestion Mr.Whitebird