<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:
    bool isIdealPermutation(vector<int>& nums) {
        int n = nums.size();
        long long local = 0, global = 0;

        for (int i = 1; i < n; ++i){
            if (nums[i-1] > nums[i]){
                ++local;
            }
        }

        global = merge_sort_and_count(nums);
        return (local == global);
    }
private:
    long long merge_and_count(vector<int>& nums, int left, int mid, int right){
        long long global = 0;
        int left_size = mid - left + 1;
        int right_size = right - mid;
        vector<int> left_arr(left_size);
        vector<int> right_arr(right_size);

        for (int i = 0; i < left_size; ++i){
            left_arr[i] = nums[left + i];
        }
        for (int i = 0; i < right_size; ++i){
            right_arr[i] = nums[mid + 1 + i];
        }

        int i = 0;
        int j = 0;
        int k = left;

        while (i < left_size && j < right_size) {
            if (left_arr[i] <= right_arr[j]) {
                nums[k++] = left_arr[i];
                i++;
            }else{
                nums[k++] = right_arr[j];
                j++;
                global += left_size - i;
            }
        }
        while (i < left_size) {
            nums[k++] = left_arr[i++];
        }
        while (j < right_size) {
            nums[k++] = right_arr[j++];
        }
        return global;
    }
    int merge_sort_and_count(vector<int>& nums){
        long long global = 0;
        int n = nums.size();
        for (int subsize = 1;  subsize < n; subsize *= 2){
            for (int left = 0; left < n; left += 2 * subsize){
                int mid = min(left + subsize - 1, n - 1);
                int right = min(left + 2 * subsize - 1, n - 1);
                global += merge_and_count(nums, left, mid, right);
            }
        }
        return global;
    }
};
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