Count of Smaller Numbers After Self
<p>Given an integer array <code>nums</code>, return<em> an integer array </em><code>counts</code><em> where </em><code>counts[i]</code><em> is the number of smaller elements to the right of </em><code>nums[i]</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [5,2,6,1]
<strong>Output:</strong> [2,1,1,0]
<strong>Explanation:</strong>
To the right of 5 there are <b>2</b> smaller elements (2 and 1).
To the right of 2 there is only <b>1</b> smaller element (1).
To the right of 6 there is <b>1</b> smaller element (1).
To the right of 1 there is <b>0</b> smaller element.
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1]
<strong>Output:</strong> [0]
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,-1]
<strong>Output:</strong> [0,0]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
Binary Indexed Tree; Segment Tree; Array; Binary Search; Divide and Conquer; Ordered Set; Merge Sort
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= this.lowbit(x)
	}
	return s
}

func countSmaller(nums []int) []int {
	s := make(map[int]bool)
	for _, v := range nums {
		s[v] = true
	}
	var alls []int
	for v := range s {
		alls = append(alls, v)
	}
	sort.Ints(alls)
	m := make(map[int]int)
	for i, v := range alls {
		m[v] = i + 1
	}
	ans := make([]int, len(nums))
	tree := newBinaryIndexedTree(len(alls))
	for i := len(nums) - 1; i >= 0; i-- {
		x := m[nums[i]]
		tree.update(x, 1)
		ans[i] = tree.query(x - 1)
	}
	return ans
}