<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:
// Recursive function with memoization to solve the problem
int solve(int num, unordered_map<int, int>& freq, unordered_map<int, int>& dp) {
    // Base case
    if (num == 0) return 0;
    if (num == 1) return freq[1];

    // If already computed, return the value from dp
    if (dp.find(num) != dp.end()) return dp[num];

    // Two choices: either take 'num' or skip 'num'
    int take = num * freq[num] + solve(num - 2, freq, dp); // Take num, skip num-1
    int skip = solve(num - 1, freq, dp); // Skip num

    // Store the result in dp and return the maximum of the two choices
    dp[num] = max(take, skip);
    return dp[num];
}


    int deleteAndEarn(vector<int>& nums) {
        unordered_map<int, int> freq;
    unordered_map<int, int> dp;

    // Create frequency map
    for (int num : nums) {
        freq[num]++;
    }

    int max_num = *max_element(nums.begin(), nums.end());

    // Call the recursive function starting from the largest number
    return solve(max_num, freq, dp);
    
        
    }
};
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