<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 merge(int low, int high, int mid, long long& count, vector<int>& nums) {
        vector<int> temp(high - low + 1);
        int left = low, right = mid + 1, k = 0;

        // Count inversions while merging
        while (left <= mid && right <= high) {
            if (nums[left] <= nums[right]) {
                temp[k++] = nums[left++];
            } else {
                temp[k++] = nums[right++];
                count += (mid - left + 1); // Count inversions
            }
        }

        // Copy remaining elements
        while (left <= mid) {
            temp[k++] = nums[left++];
        }
        while (right <= high) {
            temp[k++] = nums[right++];
        }

        // Copy sorted elements back to the original array
        for (int i = 0; i < temp.size(); ++i) {
            nums[low + i] = temp[i];
        }
    }

    void mergesort(int low, int high, vector<int>& nums, long long& count) {
        if (low >= high) return;

        int mid = low + (high - low) / 2;
        mergesort(low, mid, nums, count);
        mergesort(mid + 1, high, nums, count);
        merge(low, high, mid, count, nums);
    }

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

        vector<int> temp = nums;
        long long count = 0;
        mergesort(0, nums.size() - 1, temp, count);

        // Check if the number of inversions is at most 1
        return local == count;
    }
};

152603