<p>Given an integer array <code>nums</code> and an integer <code>k</code>, find three non-overlapping subarrays of length <code>k</code> with maximum sum and return them.</p>

<p>Return the result as a list of indices representing the starting position of each interval (<strong>0-indexed</strong>). If there are multiple answers, return the lexicographically smallest one.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,2,6,7,5,1], k = 2
<strong>Output:</strong> [0,3,5]
<strong>Explanation:</strong> Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,2,1,2,1,2,1], k = 2
<strong>Output:</strong> [0,2,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;&nbsp;2<sup>16</sup></code></li>
	<li><code>1 &lt;= k &lt;= floor(nums.length / 3)</code></li>
</ul>

class Solution {
    using LL = long long;
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        vector<LL> leftMax(n, LLONG_MIN/2);
        vector<LL> rightMax(n, LLONG_MIN/2);
        vector<int> leftIdx(n);
        vector<int> rightIdx(n);
        int idx = 0;
        LL sum = 0;
        LL maxVal = 0;
        for(int i = 0; i < n; i++){
            if(i < k-1){
                sum += (LL)nums[i];
            }
            else {
                sum += (LL)nums[i];
                if(sum > maxVal){
                    idx = i-k+1;
                    maxVal = sum;
                }
                leftMax[i] = maxVal;
                leftIdx[i] = idx;
                sum -= nums[i-k+1];
            }
        }

        // x x x x x k = 2
        maxVal = 0;
        sum = 0;
        for(int i = n-1; i >= 0; i--){
            if(i >= n-k+1){
                sum += (LL)nums[i];
            }
            else {
                sum += nums[i];
                if(sum >= maxVal){
                    idx = i;
                    maxVal = sum;
                }
                rightMax[i] = maxVal;
                rightIdx[i] = idx;
                sum -= nums[i+k-1];
            }
        }
        int idx1 = 0;
        int idx2 = 0;
        int idx3 = 0;
        maxVal = 0;
        // x x x x x x k = 2
        sum = 0;
        int i = k;
        for(int j = k; j < n-k; j++){
            sum += (LL)nums[j];
            if(j-i+1 == k){
                if(maxVal < sum + leftMax[i-1] + rightMax[j+1]){
                    maxVal = leftMax[i-1] + rightMax[j+1] + sum;
                    idx1 = leftIdx[i-1];
                    idx2 = i;
                    idx3 = rightIdx[j+1];
                }
                sum -= nums[i];
                i++;
            }
        }

        return {idx1, idx2, idx3};
    }
};
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