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
function maxTotalReward(rewardValues: number[]): number {
    rewardValues.sort((a, b) => a - b);
    const search = (x: number): number => {
        let [l, r] = [0, rewardValues.length];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (rewardValues[mid] > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const f: number[] = Array(rewardValues.at(-1)! << 1).fill(-1);
    const dfs = (x: number): number => {
        if (f[x] !== -1) {
            return f[x];
        }
        let ans = 0;
        for (let i = search(x); i < rewardValues.length; ++i) {
            ans = Math.max(ans, rewardValues[i] + dfs(x + rewardValues[i]));
        }
        return (f[x] = ans);
    };
    return dfs(0);
}
