<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>
typedef long long LL;
class Solution {
    class SegTreeNode {
        public:
        SegTreeNode* left;
        SegTreeNode* right;
        int start, end;
        int info;
        SegTreeNode(int a, int b) : start(a), end(b), info(0), left(NULL), right(NULL) {}
    };
    void init(SegTreeNode* node, int a, int b) {
        if(a == b) {
            node->info = 0;
            return;
        }
        int mid = (a + b) / 2;
        if(node->left == NULL) {
            node->left = new SegTreeNode(a, mid);
            node->right = new SegTreeNode(mid+1, b);
            init(node->left, a, mid);
            init(node->right, mid+1, b);
        }
        node->info = node->left->info + node->right->info;
    }
    void updateSingle(SegTreeNode* node, int id, int val) {
        if(id < node->start || id > node->end) return;
        if(node->start == node->end) {
            node->info += val;
            return;
        }
        updateSingle(node->left, id, val);
        updateSingle(node->right, id, val);
        node->info = node->left->info + node->right->info;
    }
    int queryRange(SegTreeNode* node, int a, int b) {
        if(b < node->start || a > node->end)
            return 0;
        if(a <= node->start && b >= node->end)
            return node->info;
        return queryRange(node->left, a, b) + queryRange(node->right, a, b);
    }
    LL M = 1e9 + 7;


public:
    int createSortedArray(vector<int>& instructions) {
        set<int> Set(instructions.begin(), instructions.end());
        unordered_map<int, int> num2Idx;
        int i = 0;
        for(auto x : Set) {
            num2Idx[x] = i;
            i++;
        }
        int n = num2Idx.size();
        SegTreeNode* root = new SegTreeNode(0, n-1);
        init(root, 0, n-1);
        long ret = 0;
        for(auto x : instructions) {
            long a = queryRange(root, 0, num2Idx[x] - 1);
            long b = queryRange(root, num2Idx[x] + 1, n - 1);
            ret += min(a, b);
            ret %= M;
            updateSingle(root, num2Idx[x], 1);
        }
        return ret;
    }
};
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