<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> dp1(n,0);
        vector<int> dp2(n,0);

        int tillithksum=0;
        for(int i=0,j=0;i<=n-k;i++){
            if(i<k){
                tillithksum+=nums[i];
                dp1[i]=tillithksum;
            }
            else{
                tillithksum+=nums[i];
                tillithksum-=nums[j];
                dp1[i]=max(dp1[i-1],tillithksum);
                j++;
            }
        }
        int fromithksum=0;
        vector<int> suffixsum(n,0);
        for(int i=n-1,j=n-1;i>=0;i--){
            if(i>=n-k){
                fromithksum+=nums[i];
                dp2[i]=fromithksum;
                suffixsum[i]=fromithksum;
            }
            else{
                fromithksum+=nums[i];
                fromithksum-=nums[j];
                suffixsum[i]=fromithksum;
                dp2[i]=max(i+1<n?dp2[i+1]:0,fromithksum);
                j--;
            }
        }

        

        int Istsubsum=0;
        int IIndsubsum=0;
        int IIIrdsubsum=0;
        int maxsum=0;
        for(int i=k;i<=n-2*k;i++){
            int first = dp1[i-1];
            int scnd=suffixsum[i];
            int third = dp2[i+k];
            int sum = first+scnd+third;
            if(sum>maxsum){
                maxsum=sum;
                Istsubsum = first;
                IIndsubsum = scnd;
                IIIrdsubsum = third;
            }
        }

        vector<int> ans(3);
        for(int i=0;i<n;i++){
            if(dp1[i]==Istsubsum){
                ans[0]=i-k+1;break;
            }
        }
        for(int i=ans[0]+k;i<n-k;i++){
            if(suffixsum[i]==IIndsubsum){
                ans[1]=i;break;
            }
        }
        for(int i=ans[1]+k;i<=n-k;i++){
            if(suffixsum[i]==IIIrdsubsum){
                ans[2]=i;break;
            }
        }

        return ans;
    }
};
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