<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;
    vector<pair<int,int>> index; // vector of pairs (value, index)
    vector<int> lesser, greater; // lesser[i] = number of values in the left of i that are less than i
                                 // greater[i] = number of values in the right of i that are greater than i
    
    void merge(int left, int mid, int right) {
        // calculate lesser and greater for each element in the right subarray
        for(int i = mid + 1, j = left, k = left; i <= right; i++) {
            while(j <= mid and index[j].first < index[i].first) j++;
            while(k <= mid and index[k].first <= index[i].first) k++;
            lesser[index[i].second] += j - left;
            greater[index[i].second] += mid - k + 1;
        }

        // merge the two sorted subarrays
        vector<pair<int,int>> temp(right - left + 1);
        int p = left, q = mid + 1, r = 0;
        while (p <= mid and q <= right) 
            if(index[p].first <= index[q].first)
                temp[r++] = index[p++];
            else
                temp[r++] = index[q++];
        while (p <= mid) 
            temp[r++] = index[p++];
        while (q <= right)
            temp[r++] = index[q++];
        
        // copy the merged subarray back to the original array
        move(temp.begin(), temp.end(), index.begin() + left);
    }

    void mergeSort(int left, int right) {
        if(left >= right)
            return;
        int mid = left + (right - left) / 2;
        mergeSort(left, mid);
        mergeSort(mid + 1, right);
        merge(left, mid, right);
    }

    int createSortedArray(vector<int>& instructions) {
        int n = instructions.size();
        index.resize(n);
        lesser.resize(n);
        greater.resize(n);

        // initialize index vector with pairs (value, index)
        for (int i = 0; i < n; i++)
            index[i] = {instructions[i], i};
        
        // sort the index vector using merge sort and calculate lesser and greater for each element
        mergeSort(0, n - 1);
        
        int ans = 0;
        for (int i = 0; i < n; i++) {
            // calculate the minimum of the two values and add it to the answer
            ans += min(lesser[i], greater[i]);
            ans %= mod;
        }
        return ans;
    }
};
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