<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 {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
       int n=nums.size();
        vector<int> left(n,0);
        vector<int> right(n,0);
        int maxSum=0;
        int leftSum=0;
        int rightSum=0;
        int curr=0;
        int maxLeft=0;
        for(int i=0;i<k;i++) curr+=nums[i];
        left[k-1]=max(curr,left[k-1]);
        maxLeft=curr;
        for(int i=k;i<n;i++){
            curr-=nums[i-k];
            curr+=nums[i];
            maxLeft=max(maxLeft,curr);
            left[i]=maxLeft;
            
        }
        
        curr=0;
        for(int i=n-k;i<n;i++) curr+=nums[i];
        int maxRight=curr;
        right[n-k]=curr;
        for(int i=n-k-1;i>=0;i--){
            curr-=nums[i+k];
            curr+=nums[i];
            maxRight=max(maxRight,curr);
            right[i]=maxRight;
        }
        
        curr=0;
        int ans=0;
        
        for(int i=k;i<2*k;i++){
            curr+=nums[i];
        }
        
        ans=max(ans,curr+left[k-1]+right[2*k]);
     
        int midIdx=k;
        int l=left[k-1];
        int r=right[2*k];
        for(int i=2*k;i<n and i+k<n;i++){
            curr+=nums[i];
            curr-=nums[i-k];
            if(ans<curr+left[i-k]+right[i+1]){
                midIdx=i-k+1;
                ans=left[i-k]+right[i+1]+curr;
                l=left[i-k];
                r=right[i+1];
            }
            
        }
     
        curr=0;
        vector<int> temp;
        for(int i=0;i<k;i++){
            curr+=nums[i];
        }
        if(curr==l) temp.push_back(0);
        else {
            for(int i=k;i<midIdx;i++){
                curr-=nums[i-k];
                curr+=nums[i];
                if(curr==l) {
                    temp.push_back(i-k+1);
                    break;
                }
            }
        }
        temp.push_back(midIdx);
        
        curr=0;
        for(int i=midIdx+k;i<midIdx+2*k;i++){
            curr+=nums[i];
        }
     
        if(curr==r){
            temp.push_back(midIdx+k);
        }
        else {
            for(int i=midIdx+2*k;i<n;i++){
                curr-=nums[i-k];
                curr+=nums[i];
                if(curr==r){
                    temp.push_back(i-k+1);
                    break;
                }
            }
        }
        
        
        
        return temp;
        
    }
};
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