<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>

class Solution {
public:
      vector<int> temp;

void merge(int l,int r,vector<int>& nums,long long& global){
    temp.clear();
      int mid=l+(r-l)/2;
      int i=l;
      int j=mid+1;

      while(i<=mid && j<=r){
        if(nums[i]>nums[j]){
            global+=(mid-i+1);
            temp.push_back(nums[j]);
            j++;
        }else{
            temp.push_back(nums[i]);
            i++;
        }
      }
      while(i<=mid){
         
        temp.push_back(nums[i]);
            i++;
      }
      while(j<=r){
         temp.push_back(nums[j]);
            j++;
      }
     int m=0;
      for(int k=l;k<=r;k++){
        nums[k]=temp[m];
        m++;
      }

}
void mergeCount(int l,int r,vector<int>& nums,long long& global){
      if(l>=r) return;
      int mid=l+(r-l)/2;
      mergeCount(l,mid,nums,global);
      mergeCount(mid+1,r,nums,global);

      merge(l,r,nums,global);

}
    bool isIdealPermutation(vector<int>& nums) {
         int local=0;
         int n=nums.size();
         for(int i=0;i<nums.size()-1;i++){
            if(nums[i]>nums[i+1]) local++;
         }

         long long global=0;
         mergeCount(0,n-1,nums,global);
         return global==local;
    }
};
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