<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:
    void inv(vector<int>&nums,int l,int r,long int &g){
        if(l>=r){return;}
        if((r-l)==1){
            if(nums[l]>nums[r]){swap(nums[l],nums[r]);g+=1;}
            return;
        }
        int m=l+(r-l)/2;
        if(l<m){inv(nums,l,m,g);}
        if(m+1<r){inv(nums,m+1,r,g);}
        int i=l,j=m+1,s=0;
        vector<int>bf((r-l+1));
        while(i<=m && j<=r){
            if(nums[i]>nums[j]){
                g+=(m-i+1);
                bf[s]=nums[j];j++;s++;
            }
            else{
                bf[s]=nums[i];i++;s++;
            }
        }
        while(i<=m){bf[s]=nums[i];i++;s++;}
        while(j<=r){bf[s]=nums[j];j++;s++;}
        s=0;
        for(int k=l;k<=r;k++){
            nums[k]=bf[s];s++;
        }
    }
    bool isIdealPermutation(vector<int>& nums) {
        long int l=0,n=nums.size();
        for(int i=0;i<n-1;i++){if(nums[i]>nums[i+1]){l+=1;}}
        long int g=0;
        inv(nums,0,n-1,g);
        //for(int i=0;i<n;i++){cout<<nums[i]<<" ";}
        cout<<g<<" "<<l<<" "<<n;
        return g==l;
    }
};
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