<p>You are given an integer array <code>nums</code> of length <code>n</code> which represents a permutation of all the integers in the range <code>[0, n - 1]</code>.</p>

<p>The number of <strong>global inversions</strong> is the number of the different pairs <code>(i, j)</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt; n</code></li>
	<li><code>nums[i] &gt; nums[j]</code></li>
</ul>

<p>The number of <strong>local inversions</strong> is the number of indices <code>i</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; n - 1</code></li>
	<li><code>nums[i] &gt; nums[i + 1]</code></li>
</ul>

<p>Return <code>true</code> <em>if the number of <strong>global inversions</strong> is equal to the number of <strong>local inversions</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,0,2]
<strong>Output:</strong> true
<strong>Explanation:</strong> There is 1 global inversion and 1 local inversion.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,0]
<strong>Output:</strong> false
<strong>Explanation:</strong> There are 2 global inversions and 1 local inversion.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt; n</code></li>
	<li>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
	<li><code>nums</code> is a permutation of all the numbers in the range <code>[0, n - 1]</code>.</li>
</ul>

#define N 100000


class Solution {
public:

    long global_invs;
    long arr[N];
    long A[N];

    void merge(long l1,long r1,long l2,long r2){
        long i=l1,j=l2,k=l1;
        while(i<=r1 || j<=r2){
            if(i>r1){
                arr[k]=A[j];
                j++;k++;
                continue;
            }
            if(j>r2){
                arr[k]=A[i];
                i++;k++;
                continue;
            }
            if(A[i]<A[j]){
                arr[k]=A[i];
                i++;k++;
            }
            else{
                arr[k]=A[j];
                j++;k++;
                global_invs += r1-i+1;
            }

        }
        for(long i = l1;i<=r2;i++)A[i]=arr[i];
    }

    void sort(long left,long right){
        if(left==right)return;
        long mid =(left + right)/2;
        sort(left,mid);
        sort(mid+1,right);
        merge(left,mid,mid+1,right);
    }



    bool isIdealPermutation(vector<int>& nums) {
        long local_invs = 0;
        for(long i = 0 ; i < nums.size() ; i++){
            A[i] = nums[i];
            if(i<nums.size()-1)local_invs += nums[i]>nums[i+1]?1:0;
        }
        global_invs = 0;
        sort(0,nums.size()-1);
        return local_invs == global_invs;
    }
};
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