<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>

# Whill always include max number m
# Best solutiom is 2m - 1
# Sort unique numbers
# Try starting at each number in this list
# Use function that starts with a current sum, and position in sorted list
#   Solves problem recursively and memoizes
class Solution:

  def __init__(self):
    self.arr = []
    self.cache = {}

  def dp(self, pos: int, reward: int) -> int:
    if pos >= len(self.arr):
      return 0

    if (pos, reward) in self.cache:
      return self.cache[(pos, reward)]

    if self.arr[pos] <= reward:
      r = self.dp(pos + 1, reward)
    else:
      r = max(
        self.arr[pos] + self.dp(pos + 1, reward + self.arr[pos]),
        self.dp(pos + 1, reward)
      )

    self.cache[(pos, reward)] = r
    return r

  def maxTotalReward(self, arr: List[int]) -> int:
    self.arr = sorted(set(arr))
    return self.dp(0, 0)


class SolutionY:
    def maxTotalReward(self, rewardValues: List[int]) -> int:
        s = set(rewardValues)
        l = list(s)
        l.sort()
        res = 0
        memo = {}
        for i in range(len(l)):
            res = max(res, self.m(l, l[i], i,  memo))
        return res

    def m(self, sv, cs, i, memo):
        if cs in memo:
            return memo[cs]
        res = cs
        for j in range(i + 1, len(sv)):
            if cs < sv[j]:
                res = max(res, self.m(sv, cs + sv[j], j, memo))
        memo[cs] = res
#        print(f"sv {sv} cs {cs} i {i} res {res}")
        return res
            
        
        
        
        
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