<p>You have a water dispenser that can dispense cold, warm, and hot water. Every second, you can either fill up <code>2</code> cups with <strong>different</strong> types of water, or <code>1</code> cup of any type of water.</p>

<p>You are given a <strong>0-indexed</strong> integer array <code>amount</code> of length <code>3</code> where <code>amount[0]</code>, <code>amount[1]</code>, and <code>amount[2]</code> denote the number of cold, warm, and hot water cups you need to fill respectively. Return <em>the <strong>minimum</strong> number of seconds needed to fill up all the cups</em>.</p>

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

<pre>
<strong>Input:</strong> amount = [1,4,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One way to fill up the cups is:
Second 1: Fill up a cold cup and a warm cup.
Second 2: Fill up a warm cup and a hot cup.
Second 3: Fill up a warm cup and a hot cup.
Second 4: Fill up a warm cup.
It can be proven that 4 is the minimum number of seconds needed.
</pre>

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

<pre>
<strong>Input:</strong> amount = [5,4,4]
<strong>Output:</strong> 7
<strong>Explanation:</strong> One way to fill up the cups is:
Second 1: Fill up a cold cup, and a hot cup.
Second 2: Fill up a cold cup, and a warm cup.
Second 3: Fill up a cold cup, and a warm cup.
Second 4: Fill up a warm cup, and a hot cup.
Second 5: Fill up a cold cup, and a hot cup.
Second 6: Fill up a cold cup, and a warm cup.
Second 7: Fill up a hot cup.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> amount = [5,0,0]
<strong>Output:</strong> 5
<strong>Explanation:</strong> Every second, we fill up a cold cup.
</pre>

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

<ul>
	<li><code>amount.length == 3</code></li>
	<li><code>0 &lt;= amount[i] &lt;= 100</code></li>
</ul>

# # 12.35/12.45/0/0/0
# greedy solution:
# pop 2 cups off (if there are different types of water left to pour)
# or n steps for n remaining cups of a single type of water

# add in 3 elements to a heap
# pop first 2, minus 1 each 
# if nonzero, reinsert
# continue until heap is empty or only 1 node is left

import heapq
class Solution:
    def fillCups(self, amount: List[int]) -> int:
        cupheap = [-cups for cups in amount if cups > 0]
        heapq.heapify(cupheap)
        totalSteps = 0
        while len(cupheap)>0:
            if len(cupheap) == 1:
                totalSteps += -cupheap[0]
                break
            cups1 = heapq.heappop(cupheap)
            cups2 = heapq.heappop(cupheap)
            cups1 +=1
            cups2 +=1
            totalSteps +=1
            if cups1:
                heapq.heappush(cupheap, cups1)
            if cups2:
                heapq.heappush(cupheap, cups2)
            print(cupheap)
        return totalSteps
45