<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>
class SegmentTree {
    vector<int> tree;
    int maxVal;

public:
    SegmentTree(int maxVal) : maxVal(maxVal) {
        tree.resize(4 * maxVal, 0);
    }

    void update(int node, int start, int end, int idx, int value) {
        if (start == end) {
            // Leaf node: update the frequency
            tree[node] += value;
        } else {
            int mid = (start + end) / 2;
            if (idx <= mid) {
                update(2 * node, start, mid, idx, value);
            } else {
                update(2 * node + 1, mid + 1, end, idx, value);
            }
            // Internal node: sum of left and right children
            tree[node] = tree[2 * node] + tree[2 * node + 1];
        }
    }

    int query(int node, int start, int end, int l, int r) {
        if (r < start || end < l) {
            // Completely outside the range
            return 0;
        }
        if (l <= start && end <= r) {
            // Completely inside the range
            return tree[node];
        }
        // Partially inside and outside
        int mid = (start + end) / 2;
        int left_query = query(2 * node, start, mid, l, r);
        int right_query = query(2 * node + 1, mid + 1, end, l, r);
        return left_query + right_query;
    }

    void update(int idx, int value) {
        update(1, 0, maxVal, idx, value);
    }

    int query(int l, int r) {
        return query(1, 0, maxVal, l, r);
    }
};

class Solution {
public:
    int createSortedArray(vector<int>& instructions) {
        int mod = 1e9 + 7;
        int maxVal = 1e5;
        SegmentTree segTree(maxVal);
        long long cost = 0;

        for (int i = 0; i < instructions.size(); ++i) {
            int num = instructions[i];
            int lessCount = segTree.query(0, num - 1);
            int greaterCount = segTree.query(num + 1, maxVal);
            cost += min(lessCount, greaterCount);
            cost %= mod;
            segTree.update(num, 1);
        }

        return cost;
    }
};

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