Can u provid me the c++ code with the most optimis...

Yes. The cleanest solution is O(N²) using prefix sums. The important part is that an overlapping element must be counted once with its sign reversed.

cpp
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

int solution(vector<int> &A, int K, int L) {
    int N = A.size();
    vector<int> pre(N + 1, 0);

    for (int i = 0; i < N; ++i)
        pre[i + 1] = pre[i] + A[i];

    auto sum = [&](int l, int r) {
        return pre[r] - pre[l];
    };

    int ans = INT_MIN;

    for (int i = 0; i + K <= N; ++i) {
        for (int j = 0; j + L <= N; ++j) {

            int total = sum(i, i + K) + sum(j, j + L);

            int left = max(i, j);
            int right = min(i + K, j + L);

            if (left < right)
                total -= 3 * sum(left, right);

            ans = max(ans, total);
        }
    }

    return ans;
}

Why -3 * overlap?

Initially, sum(K fragment) + sum(L fragment) counts every overlapping value twice:

2 × A[i]

But the problem wants it counted once with the opposite sign:

-A[i]

So:

2A[i] - 3A[i] = -A[i]

For example, if the overlapping value is -4:

2(-4) - 3(-4) = -8 + 12 = 4

which correctly changes -4 into +4.

This handles positive/negative values, partial overlap, complete overlap, K=L, K=N/L=N, and the examples shown. Complexity is O(N²) time and O(N) memory.