<p>Given an integer array <code>nums</code>, return <em>the number of <strong>reverse pairs</strong> in the array</em>.</p>

<p>A <strong>reverse pair</strong> is a pair <code>(i, j)</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt; nums.length</code> and</li>
	<li><code>nums[i] &gt; 2 * nums[j]</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> nums = [1,3,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 3 &gt; 2 * 1
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,3,5,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 4, nums[4] = 1, 4 &gt; 2 * 1
(2, 4) --&gt; nums[2] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 5, nums[4] = 1, 5 &gt; 2 * 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        """
        Divide and Conquer approach:
            - divide array into two halves
            - sort first half, and get counting for second half
            - using binary search on first half, find number of pairs i in first half, and 
                j in second half such that the condition holds
            - apply above steps on each half (unsorted)
            - final answer is total of each subparts
        """

        def _numsGreaterThanTarget(array, target):
            # binary search, duplicates possible
            left = 0
            right = len(array)-1
            while left<right:
                mid = (left+right)//2
                # [4, 5] target=4
                if array[mid]<=target:
                    left = mid+1
                else:
                    right = mid
            return len(array)-left if array[left]>target else len(array)-left-1

        def _reversePairsBothHalves(firsthalf, secondhalf):
            # firsthalf: sortednums
            # secondhalf: counter of each occurences
            total = 0
            for key in secondhalf:
                total += secondhalf[key]*_numsGreaterThanTarget(firsthalf, 2*key)
            return total

        def _reversePairs(start, end):
            if end-start<1:
                return 0
            # divide and conquer
            mid = (start+end)//2
            # edge-cases: [0, 1], and [0, 1, 2]
            firsthalf = sorted(nums[start:mid+1])
            secondhalf = collections.Counter(nums[mid+1:end+1])
            total = _reversePairsBothHalves(firsthalf, secondhalf)
            return total + _reversePairs(start, mid) + _reversePairs(mid+1, end)

        return _reversePairs(0, len(nums)-1)
# TC: O(N log(N))
# SC: O(N)
31600