<p>You are given an integer array <code>nums</code>. You want to maximize the number of points you get by performing the following operation any number of times:</p>

<ul>
	<li>Pick any <code>nums[i]</code> and delete it to earn <code>nums[i]</code> points. Afterwards, you must delete <b>every</b> element equal to <code>nums[i] - 1</code> and <strong>every</strong> element equal to <code>nums[i] + 1</code>.</li>
</ul>

<p>Return <em>the <strong>maximum number of points</strong> you can earn by applying the above operation some number of times</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,4,2]
<strong>Output:</strong> 6
<strong>Explanation:</strong> You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,3,3,3,4]
<strong>Output:</strong> 9
<strong>Explanation:</strong> You can perform the following operations:
- Delete a 3 to earn 3 points. All 2&#39;s and 4&#39;s are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.</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;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public: 
    // AC - AKS - solved myself - 2D DP Tabulation myself
    int deleteAndEarn(vector<int>& nums) {

        map<int,int>mp;
        int Mx = 0;
        for(auto x : nums) {
            mp[x]++;
            Mx = max(Mx,x);
        }

        // dp[i][0] ==> max sum when we don't take element i
        // dp[i][1] ==> max sum when we take element i
        vector<vector<int>>dp(Mx+1,vector<int>(2,0));
        int ans = 0;
        for(int i = 1; i <= Mx; i++){
            // If i is not present in map
            // then just don't take it
            if(mp.find(i) == mp.end()){
                // don't take current element
                dp[i][0] = max(dp[i-1][1], dp[i-1][0]);
                continue;
            }

            // pick current element
            // We should not have taken previous i-1
            dp[i][1] = i*mp[i] + dp[i-1][0];

            // don't take current element
            // We could have taken or not taken previous i-1
            dp[i][0] = max(dp[i-1][1], dp[i-1][0]);

            ans = max({ans,dp[i][1], dp[i][0]});
        }

        return ans;
    }
};
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