<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 constrainedSubsetSum(vector<int>& nums, int k) {
        int i=0,j=0;
        vector<long>dp(nums.size(),0);
        map<int,int>mp;
        while(j<nums.size())
        {
            while(j>=i&&j-i>k)
            {
                mp[dp[i]]--;
                if(mp[dp[i]]==0)mp.erase(dp[i]);
                 
                i++;
             
            }
            if(mp.size()>0)
           { 
            dp[j]=max(0,mp.rbegin()->first)+nums[j];
            mp[dp[j]]++;
            }
            else
            {
                dp[j]=nums[j];
                mp[dp[j]]++;
            }
            j++;
        }
        // int mine=0;
        // for(auto it:nums)
        // {
        //     if(it>=0)mine=1;
        // }
        for(auto it:dp)cout<<it<<" ";
        // if(!mine)return *max_element(nums.begin(),nums.end());
        return *max_element(dp.begin(),dp.end());
    }
};
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