<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) {
        // multiset<int> st;
        map<int,int> mp;
        int sum = 0;
        int j = 0;
        // st.insert(nums[0]);
        mp[nums[0]]++;
        vector<int> v;
        int mn = -1e9;
        v.push_back(nums[0]);
        mn = nums[0];
        for(int i = 1;i < nums.size();i++){
            if(i > k){
                // st.erase(st.find(v[j]));
                mp[v[j]]--;
                if(mp[v[j]] == 0){
                    mp.erase(v[j]);
                }
                j++;
            }
            auto en = mp.end();
            en--;
            if(en->first < 0){
                // st.insert(nums[i]);
                mp[nums[i]]++;
                v.push_back(nums[i]);
            }
            else{
                mp[(nums[i] + en->first)]++;
                // st.insert(nums[i] + en->first);
                v.push_back(nums[i] + en->first);
            }
            mn = max(mn,v[i]);

        }
        // for(int i = 0;i < nums.size();i++){
        //     // cout<<v[i]<<" ";
        //     mn = max(mn,v[i]);
        // }
        return mn;
        
    }
};
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