<p>Your country has an infinite number of lakes. Initially, all the lakes are empty, but when it rains over the <code>nth</code> lake, the <code>nth</code> lake becomes full of water. If it rains over a lake that is <strong>full of water</strong>, there will be a <strong>flood</strong>. Your goal is to avoid floods in any lake.</p>

<p>Given an integer array <code>rains</code> where:</p>

<ul>
	<li><code>rains[i] &gt; 0</code> means there will be rains over the <code>rains[i]</code> lake.</li>
	<li><code>rains[i] == 0</code> means there are no rains this day and you can choose <strong>one lake</strong> this day and <strong>dry it</strong>.</li>
</ul>

<p>Return <em>an array <code>ans</code></em> where:</p>

<ul>
	<li><code>ans.length == rains.length</code></li>
	<li><code>ans[i] == -1</code> if <code>rains[i] &gt; 0</code>.</li>
	<li><code>ans[i]</code> is the lake you choose to dry in the <code>ith</code> day if <code>rains[i] == 0</code>.</li>
</ul>

<p>If there are multiple valid answers return <strong>any</strong> of them. If it is impossible to avoid flood return <strong>an empty array</strong>.</p>

<p>Notice that if you chose to dry a full lake, it becomes empty, but if you chose to dry an empty lake, nothing changes.</p>

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

<pre>
<strong>Input:</strong> rains = [1,2,3,4]
<strong>Output:</strong> [-1,-1,-1,-1]
<strong>Explanation:</strong> After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day full lakes are [1,2,3]
After the fourth day full lakes are [1,2,3,4]
There&#39;s no day to dry any lake and there is no flood in any lake.
</pre>

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

<pre>
<strong>Input:</strong> rains = [1,2,0,0,2,1]
<strong>Output:</strong> [-1,-1,2,1,-1,-1]
<strong>Explanation:</strong> After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day, we dry lake 2. Full lakes are [1]
After the fourth day, we dry lake 1. There is no full lakes.
After the fifth day, full lakes are [2].
After the sixth day, full lakes are [1,2].
It is easy that this scenario is flood-free. [-1,-1,1,2,-1,-1] is another acceptable scenario.
</pre>

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

<pre>
<strong>Input:</strong> rains = [1,2,0,1,2]
<strong>Output:</strong> []
<strong>Explanation:</strong> After the second day, full lakes are  [1,2]. We have to dry one lake in the third day.
After that, it will rain over lakes [1,2]. It&#39;s easy to prove that no matter which lake you choose to dry in the 3rd day, the other one will flood.
</pre>

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

<ul>
	<li><code>1 &lt;= rains.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= rains[i] &lt;= 10<sup>9</sup></code></li>
</ul>

from collections import defaultdict
from heapq import heappop,heappush

class Solution:
    def avoidFlood(self, rains: List[int]) -> List[int]:
        lake_rains = defaultdict(list)
        n = len(rains)

        """
        To solve this problem we must know what type of 
        lakes are infront of us and from the available lake
        we will always choose the one that is the nearest and 
        """

        # track the list of days on when it will 
        # rain for each day.
        for i in range(n):
            lake_rains[rains[i]].append(i)

        full_lakes = set()
        heap = [] # will store potential future floods
        solution = [-1]*n

        for i in range(n):
            # it will rain 
            if rains[i]:
                if rains[i] in full_lakes:
                    return [] # flood unavoidable
            
                # mark the lake as full for future iterations
                full_lakes.add(rains[i])
                # remove from lake rains because it is used to fill the lake 
                lake_rains[rains[i]].pop(0)

                if lake_rains[rains[i]]:
                    # add the next filling day to future fills
                    heappush(heap,lake_rains[rains[i]][0])

            # it will not rain
            else:
                # there are lakes to dry
                if heap:
                    closest_day = heappop(heap)
                    # dry the current lake
                    solution[i] = rains[closest_day]
                    full_lakes.remove(rains[closest_day])
                # there are no lakes to dry
                # dry any of them
                else:
                    solution[i] = 1


        return solution


        
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