<p>Given an array of integers&nbsp;<code>arr</code>, replace each element with its rank.</p>

<p>The rank represents how large the element is. The rank has the following rules:</p>

<ul>
	<li>Rank is an integer starting from 1.</li>
	<li>The larger the element, the larger the rank. If two elements are equal, their rank must be the same.</li>
	<li>Rank should be as small as possible.</li>
</ul>

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

<pre>
<strong>Input:</strong> arr = [40,10,20,30]
<strong>Output:</strong> [4,1,2,3]
<strong>Explanation</strong>: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.</pre>

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

<pre>
<strong>Input:</strong> arr = [100,100,100]
<strong>Output:</strong> [1,1,1]
<strong>Explanation</strong>: Same elements share the same rank.
</pre>

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

<pre>
<strong>Input:</strong> arr = [37,12,28,9,100,56,80,5,12]
<strong>Output:</strong> [5,3,4,2,8,6,7,1,3]
</pre>

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

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

class Solution:
    def arrayRankTransform(self, arr: List[int]) -> List[int]:
        if not arr:
            return []
        
        numToArrIdx = [(num, arrIdx) for arrIdx, num in enumerate(arr)]
        numToArrIdx.sort()
        
        rank = 1
        
        for idx, (num, arrIdx) in enumerate(numToArrIdx):
            arr[arrIdx] = rank
            # ONLY increment rank if next number isn't the same
            if idx < len(numToArrIdx) - 1 and num != numToArrIdx[idx + 1][0]:
                rank += 1
                    
        return arr
        
        
        
        
        
"""
brute force might be finding the min or duplicates, setting rank
"removing" min (and duplicates) and reprocess O(N^2)

faster sort first, but we keep idx in the sort (to reference) idx in
original arr

O(nlogn) 

in place?
empty array?

  0  1  2  3
[40,10,20,30]

(10, 1), (20, 2) (30, 3), (40, 4)
0) [40,1,20,30] | 10 != 20 => rank += 1 = 2
1) [40,1,2,30] | 20 != 30 => rank += 1 = 3 
2) [40,1,2,3] | 30 != 40 => rank += 1 = 4
3) [4,1,2,3]


 0 1 2  3  4   5  6  7 8
[5,9,12,12,28,37,56,80,100]

rank = 1
0) [37,12,28,9,100,56,80,[1],12] | 5 != 9, rank = 2
1) [37,12,28,[2],100,56,80,[1],12] 9 != 12, rank = 3
2) [37,[3],28,[2],100,56,80,[1],12] 12 == 12, rank = 3
2) [37,[3],28,[2],100,56,80,[1],[3]] 12 != 28, rank = 4
3) [37,[3],28,[2],100,56,80,[1],[3]]
"""
34400