// Time:  O(n)
// Space: O(n)

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        const auto n = nums.size();
        vector<int> accu = {0};
        for (const auto& num : nums) {
            accu.emplace_back(accu.back() + num);
        }
       
        vector<int> left_pos(n);
        for (int i = k, total = accu[k] - accu[0]; i < n; ++i) {
            if (accu[i + 1] - accu[i + 1 - k] > total) {
                left_pos[i] = i + 1 - k;
                total = accu[i + 1] - accu[i + 1 - k];
            } else { 
                left_pos[i] = left_pos[i - 1];
            }
        }
        
        vector<int> right_pos(n, n - k);
        for (int i = n - k - 1, total = accu[n] - accu[n - k]; i >= 0; --i) {
            if (accu[i + k] - accu[i] > total) {
                right_pos[i] = i;
                total = accu[i + k] - accu[i];
            } else {
                right_pos[i] = right_pos[i + 1];
            }
        }
        
        vector<int> result(3);
        for (int i = k, max_sum = 0; i <= n - 2 * k; ++i) {
            auto left = left_pos[i - 1], right = right_pos[i + k];
            auto total = (accu[i + k] - accu[i]) +
                         (accu[left + k] - accu[left]) +
                         (accu[right + k] - accu[right]);
            if (total > max_sum) {
                max_sum = total;
                result = {left, i, right};
            }
        }
        return result;
    }
};