<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 {
    unsigned int gI;

    void merge(vector<int>& a, int l, int mid, int r) {
        int i = l;
        int j = mid + 1;
        
        vector<int> b(r - l + 1);
        int k = 0;
        
        while(i <= mid && j <= r) {
            if(a[i] > a[j]) {
                b[k] = a[j];
                ++j, ++k;
                gI += mid - i + 1;
            }
            else {
                b[k] = a[i];
                ++i, ++k;
            }
        }
        while(i <= mid) b[k++] = a[i++];
        while(j <= r) b[k++] = a[j++];
        

        for(int i = 0; i < b.size(); ++i) a[l + i] = b[i];
    }
    void mergesort(vector<int>& a, int i, int j) {
        if(i >= j) return;

        int mid = (i + j) / 2;
        mergesort(a, i, mid);
        mergesort(a, mid + 1, j);
        merge(a, i, mid, j);
    }
public:
    bool isIdealPermutation(vector<int>& nums) {
        int N = nums.size();
        int lI = 0;
        for(int i = 0; i < N - 1; ++i) lI += (nums[i] > nums[i + 1]);
        
        gI = 0;
        mergesort(nums, 0, N - 1);

        return lI == gI;
    }
};
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