<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 Solution {
public:
    int MOD = 1e9 + 7;
    int createSortedArray(vector<int>& instructions) {
        int n = instructions.size();


        vector<long> cntlarger(n, 0);
        vector<int> sorted(n); // index in sorted order
        for (int i = 0; i < n; i++)
            sorted[i] = i;

        divcon(instructions, sorted, cntlarger, 0, n-1);

        long ret = 0;
        unordered_map<int, long> cnt;
        for (long i = 0; i < n; i++) {
            long smaller = i - cntlarger[i] - cnt[instructions[i]];
            long cost = min(smaller, cntlarger[i]);
            ret = (ret + cost) % MOD;
            cnt[instructions[i]]++;
        }
        return ret;

    }

    void divcon(const vector<int>& instructions, vector<int>& sorted, vector<long>& cntlarger, int l, int r) {
        if (l >= r) return;

        // divide
        int mid = (l + r) / 2;
        divcon(instructions, sorted, cntlarger, l, mid);
        divcon(instructions, sorted, cntlarger, mid+1, r);

        // conquer
        int i = l, j = mid+1;
        vector<int> tmp(r - l + 1);
        for (int k = 0; k <= (r - l); k++) {
            if (i > mid) {
                tmp[k] = sorted[j++];
            } else if (j > r) {
                tmp[k] = sorted[i++];
            } else if (instructions[sorted[i]] <= instructions[sorted[j]]) {
                tmp[k] = sorted[i++];
            } else {
                cntlarger[sorted[j]] += (mid + 1 - i);
                tmp[k] = sorted[j++];
            }
        }

        // update sorted
        for (int k = 0; k <= (r - l); k++) {
            sorted[l + k] = tmp[k];
        }
    }
};

/**
like count of smaller numbers after self,

the larger elements on the left of a number == numbers jump from left to right during stable sort
(because the same value number will remain the same order)

merge A, B to C

if A[i] <= B[j]
place A[i]
if A[i] > B[j]
place B[j], (n-i) elements jump from left of B[j] to right of B[j]


*/

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