<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> pre(n),v;
        
            int x=0;
        for(int i=0;i<n;i++){
            pre[i]=x+nums[i];
            x+=nums[i];
            cout<<pre[i]<<" ";
        }
        v.push_back(pre[k-1]);
        for(int i=k;i<n;i++){
            v.push_back(pre[i]-pre[i-k]);
        }
        cout<<endl;
        for(int i=0;i<v.size();i++){
            cout<<v[i]<<" ";
        }

            vector<int> ans;
            int sum=0;
            for(int i=k;i<v.size();i++){
                vector<int> temp;
                int tot=0;
                int ind1=0;
                int val1=0;
                for(int j=0;j<=i-k;j++){
                    if(val1<v[j]){
                        ind1=j;
                        val1=v[j];
                    }
                }
                tot+=val1;
                temp.push_back(ind1);
                tot+=v[i];
                temp.push_back(i);
                ind1=i+k;
                val1=0;
                if(ind1>=v.size())break;
                for(int j=i+k;j<v.size();j++){
                    if(v[j]>val1){
                        val1=v[j];
                        ind1=j;
                    }
                }
                tot+=v[ind1];
                temp.push_back(ind1);
                if(tot>sum){
                    ans=temp;
                    sum=tot;
                }

            }
        return ans;
        

        

    }
};
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