Maximum Total Reward Using Operations I
<p>You are given an integer array <code>rewardValues</code> of length <code>n</code>, representing the values of rewards.</p>

<p>Initially, your total reward <code>x</code> is 0, and all indices are <strong>unmarked</strong>. You are allowed to perform the following operation <strong>any</strong> number of times:</p>

<ul>
	<li>Choose an <strong>unmarked</strong> index <code>i</code> from the range <code>[0, n - 1]</code>.</li>
	<li>If <code>rewardValues[i]</code> is <strong>greater</strong> than your current total reward <code>x</code>, then add <code>rewardValues[i]</code> to <code>x</code> (i.e., <code>x = x + rewardValues[i]</code>), and <strong>mark</strong> the index <code>i</code>.</li>
</ul>

<p>Return an integer denoting the <strong>maximum </strong><em>total reward</em> you can collect by performing the operations optimally.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">rewardValues = [1,1,3,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">rewardValues = [1,6,4,3,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">11</span></p>

<p><strong>Explanation:</strong></p>

<p>Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.</p>
</div>

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

<ul>
	<li><code>1 &lt;= rewardValues.length &lt;= 2000</code></li>
	<li><code>1 &lt;= rewardValues[i] &lt;= 2000</code></li>
</ul>
Array; Dynamic Programming
class Solution {
public:
    int maxTotalReward(vector<int>& rewardValues) {
        sort(rewardValues.begin(), rewardValues.end());
        int n = rewardValues.size();
        int f[rewardValues.back() << 1];
        memset(f, -1, sizeof(f));
        function<int(int)> dfs = [&](int x) {
            if (f[x] != -1) {
                return f[x];
            }
            auto it = upper_bound(rewardValues.begin(), rewardValues.end(), x);
            int ans = 0;
            for (; it != rewardValues.end(); ++it) {
                ans = max(ans, rewardValues[it - rewardValues.begin()] + dfs(x + *it));
            }
            return f[x] = ans;
        };
        return dfs(0);
    }
};