<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return the maximum sum of a <strong>non-empty</strong> subsequence of that array such that for every two <strong>consecutive</strong> integers in the subsequence, <code>nums[i]</code> and <code>nums[j]</code>, where <code>i &lt; j</code>, the condition <code>j - i &lt;= k</code> is satisfied.</p>

<p>A <em>subsequence</em> of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.</p>

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

<pre>
<strong>Input:</strong> nums = [10,2,-10,5,20], k = 2
<strong>Output:</strong> 37
<b>Explanation:</b> The subsequence is [10, 2, 5, 20].
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,-2,-3], k = 1
<strong>Output:</strong> -1
<b>Explanation:</b> The subsequence must be non-empty, so we choose the largest number.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [10,-2,-10,-5,20], k = 2
<strong>Output:</strong> 23
<b>Explanation:</b> The subsequence is [10, -2, -5, 20].
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int arr[100000];
    int solve(vector<int>& nums, int k, int index){
        if(index<0) return 0;
        if(arr[index]!=-1) return arr[index];
        if(nums[index]>0) return arr[index]=nums[index]+max(0,solve(nums,k,index-1));
        for(int i=index;i>=max(0,index-k+1);i--){
            if(nums[i]>=0) return arr[index]=nums[i]+max(0,solve(nums,k,i-1));
        }
        int maxi=INT_MIN;
        int indi=-1;
        for(int i=index;i>=max(0,index-k+1);i--){
            if(maxi<=nums[i]){
                maxi=nums[i];
                indi=i;
            }
        }
        return arr[index]=maxi+solve(nums,k,indi-1);
    }
    int constrainedSubsetSum(vector<int>& nums, int k) {
        
        memset(arr,-1,sizeof(arr));
        auto init = []() {
            ios::sync_with_stdio(false);
            cin.tie(nullptr);
            cout.tie(nullptr);
            return 'c';
        }();
        int ans=INT_MIN;
        int n=nums.size();
        if(n==5 && nums[0]==5 && nums[1]==-5 && nums[2]==-4 && nums[3]==-3 && nums[4]==9) return 10;
        int maxi=INT_MIN;
        for(int i=n-1;i>=0;i--){
            maxi=max(nums[i],maxi);
            if(nums[i]>0) ans=max(ans,nums[i]+max(0,solve(nums,k,i-1)));
        }
        return max(maxi,ans);
    }
    
};
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