<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:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        s=0
        left=[[0,0] for i in range(0,len(nums))]
        right=[[0,0] for i in range(0,len(nums))]
        for i in range(0,len(nums)):
            if(i<k):
                s+=nums[i]
                left[i]=[s,0] 
            else:
                s+=nums[i]-nums[i-k]
                if(left[i-1][0]<s):
                    left[i][0]=s
                    left[i][1]=i-k+1 
                else:
                    left[i][0]=left[i-1][0]
                    left[i][1]=left[i-1][1]
        s=0
        for i in range(len(nums)-1,-1,-1):
            if(i>=(len(nums)-k)):
                s+=nums[i]
                right[i]=[s,len(nums)-k] 
            else:
                s+=nums[i]-nums[i+k]
                if(right[i+1][0]<=s):
                    right[i][0]=s
                    right[i][1]=i 
                else:
                    right[i][0]=right[i+1][0]
                    right[i][1]=right[i+1][1]
        s=0
        p=[] 
        for i in range(0,len(nums)):
            if(i==0):
                p.append(nums[i])
            else:
                p.append(nums[i]+p[-1])
        s1=0
        f=[]
        print(left,right)
        for i in range(k,len(nums)-(2*k)+1):
            s2=left[i-1][0]+right[i+k][0]+p[i+k-1]-p[i-1]
            if(s2>s1):
                s1=s2
                print(left[i-1],i,right[i+k])
                f=[left[i-1][1],i,right[i+k][1]]

               
        return f



        




            


        
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