<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>
void update(vector<long long> &seg_tree, int ls, int rs, int l, int r, long long pos, int addV){

    if(ls<=l && r<=rs){
        if(l!=r){
            long long mid = ((long long)l+(long long)r)/2;

            update(seg_tree, ls, rs, l, mid, 2*pos+1, addV);
            update(seg_tree, ls, rs, mid+1, r, 2*pos+2, addV);
            seg_tree[pos] = seg_tree[2*pos+1] + seg_tree[2*pos+2];

        }
        else{
            seg_tree[pos] += addV;
        }
    }
    else if(ls<=r && rs>=l){
        long long mid = (l+r)/2;
        update(seg_tree, ls, rs, l, mid, 2*pos+1, addV);
        update(seg_tree, ls, rs, mid+1, r, 2*pos+2, addV);
        seg_tree[pos] = seg_tree[2*pos+1]+seg_tree[2*pos+2];
    }else{
        return;
    }
}
long long query(vector<long long> &seg_tree, int ls, int rs, int l, int r, int pos){
    if(ls<=l && rs>=r){
        return seg_tree[pos];
    }
    else if(ls<=r && rs>=l){
        long long mid = (l+r)/2;
        long long a = query(seg_tree, ls, rs, l, mid, 2*pos+1);
        long long b = query(seg_tree, ls, rs, mid+1, r, 2*pos+2);
        return a+b;

    }
    else{
        return 0;
    }
}
class Solution {    
public:
    int createSortedArray(vector<int>& instructions) {
        const int N = 1e5; 
        const long long mod = 1e9 + 7;
        vector<long long> seg_tree(4*N+10);
        long long ans = 0;
        unordered_map<int, int> mp;
        for(int i=0; i<instructions.size(); i++){
            mp[instructions[i]]++;
            update(seg_tree, instructions[i], instructions[i], 0, N, 0, 1);
            long long a = query(seg_tree, 0, instructions[i]-1, 0, N, 0);
            long long b = i+1-mp[instructions[i]]-a;
            ans+= min(a, b)%mod;
            ans = ans%mod;
        }
        return ans;
        
    }
};
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