Create Sorted Array through Instructions
<p>Given an integer array <code>instructions</code>, you are asked to create a sorted array from the elements in <code>instructions</code>. You start with an empty container <code>nums</code>. For each element from <strong>left to right</strong> in <code>instructions</code>, insert it into <code>nums</code>. The <strong>cost</strong> of each insertion is the <b>minimum</b> of the following:</p>

<ul>

    <li>The number of elements currently in <code>nums</code> that are <strong>strictly less than</strong> <code>instructions[i]</code>.</li>

    <li>The number of elements currently in <code>nums</code> that are <strong>strictly greater than</strong> <code>instructions[i]</code>.</li>

</ul>

<p>For example, if inserting element <code>3</code> into <code>nums = [1,2,3,5]</code>, the <strong>cost</strong> of insertion is <code>min(2, 1)</code> (elements <code>1</code> and <code>2</code> are less than <code>3</code>, element <code>5</code> is greater than <code>3</code>) and <code>nums</code> will become <code>[1,2,3,3,5]</code>.</p>

<p>Return <em>the <strong>total cost</strong> to insert all elements from </em><code>instructions</code><em> into </em><code>nums</code>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code></p>

<p>&nbsp;</p>

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

<pre>

<strong>Input:</strong> instructions = [1,5,6,2]

<strong>Output:</strong> 1

<strong>Explanation:</strong> Begin with nums = [].

Insert 1 with cost min(0, 0) = 0, now nums = [1].

Insert 5 with cost min(1, 0) = 0, now nums = [1,5].

Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].

Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].

The total cost is 0 + 0 + 0 + 1 = 1.</pre>

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

<pre>

<strong>Input:</strong> instructions = [1,2,3,6,5,4]

<strong>Output:</strong> 3

<strong>Explanation:</strong> Begin with nums = [].

Insert 1 with cost min(0, 0) = 0, now nums = [1].

Insert 2 with cost min(1, 0) = 0, now nums = [1,2].

Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].

Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].

Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].

Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].

The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.

</pre>

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

<pre>

<strong>Input:</strong> instructions = [1,3,3,3,2,4,2,1,2]

<strong>Output:</strong> 4

<strong>Explanation:</strong> Begin with nums = [].

Insert 1 with cost min(0, 0) = 0, now nums = [1].

Insert 3 with cost min(1, 0) = 0, now nums = [1,3].

Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].

Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].

Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].

Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].

​​​​​​​Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].

​​​​​​​Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].

​​​​​​​Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].

The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.

</pre>

<p>&nbsp;</p>

<p><strong>Constraints:</strong></p>

<ul>

    <li><code>1 &lt;= instructions.length &lt;= 10<sup>5</sup></code></li>

    <li><code>1 &lt;= instructions[i] &lt;= 10<sup>5</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) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += x & -x
	}
}

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

func createSortedArray(instructions []int) (ans int) {
	m := slices.Max(instructions)
	tree := newBinaryIndexedTree(m)
	const mod = 1e9 + 7
	for i, x := range instructions {
		cost := min(tree.query(x-1), i-tree.query(x))
		ans = (ans + cost) % mod
		tree.update(x, 1)
	}
	return
}