<p>You are given an array <code>nums</code>. You can rotate it by a non-negative integer <code>k</code> so that the array becomes <code>[nums[k], nums[k + 1], ... nums[nums.length - 1], nums[0], nums[1], ..., nums[k-1]]</code>. Afterward, any entries that are less than or equal to their index are worth one point.</p>

<ul>
	<li>For example, if we have <code>nums = [2,4,1,3,0]</code>, and we rotate by <code>k = 2</code>, it becomes <code>[1,3,0,2,4]</code>. This is worth <code>3</code> points because <code>1 &gt; 0</code> [no points], <code>3 &gt; 1</code> [no points], <code>0 &lt;= 2</code> [one point], <code>2 &lt;= 3</code> [one point], <code>4 &lt;= 4</code> [one point].</li>
</ul>

<p>Return <em>the rotation index </em><code>k</code><em> that corresponds to the highest score we can achieve if we rotated </em><code>nums</code><em> by it</em>. If there are multiple answers, return the smallest such index <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,1,4,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Scores for each k are listed below: 
k = 0,  nums = [2,3,1,4,0],    score 2
k = 1,  nums = [3,1,4,0,2],    score 3
k = 2,  nums = [1,4,0,2,3],    score 3
k = 3,  nums = [4,0,2,3,1],    score 4
k = 4,  nums = [0,2,3,1,4],    score 3
So we should choose k = 3, which has the highest score.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,0,2,4]
<strong>Output:</strong> 0
<strong>Explanation:</strong> nums will always have 3 points no matter how it shifts.
So we will choose the smallest k, which is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt; nums.length</code></li>
</ul>

class Solution:
    def bestRotation(self, nums: List[int]) -> int:
        n = len(nums)
        change = [0] * (n + 1)  # To store net point changes at each rotation index

        # Calculate the effect of each element on the rotation score
        for i in range(n):
            # If nums[i] is placed at index i, rotation index would be (i - nums[i]) % n
            # We gain points if nums[i] <= current index after rotation
            a, b = (i - nums[i] + 1) % n, (i + 1) % n
            change[a] -= 1  # Rotation at index a decreases the score
            change[b] += 1  # Rotation at index b increases the score
            if a >= b: 
                change[0] += 1  # Adjust score for wrap-around case

        # Find the rotation index with the maximum score
        max_points, best_k, current_points = -1, 0, 0
        for k in range(n):
            current_points += change[k]
            if current_points > max_points:
                max_points = current_points
                best_k = k

        return best_k
30100