<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 {
public:
    int bestRotation(vector<int>& nums) {
        int n = nums.size();
        vector<int> nums2 = nums;
        for(int i=0;i<n;i++)nums2.push_back(nums[i]);
        for(int i=0;i<2*n;i++)nums2[i] -= i;
        int ans = 0;
        // unordered_map<int,int> mp;
        int mpt[4*n];
        memset(mpt,0,sizeof(mpt));
        int* mp = mpt + 2*n;
        for(int i=0;i<n;i++)
        {
            mp[nums2[i]] ++;
            if(nums2[i] <= 0)ans++;
        }

        int fin_ans = ans;
        int fin_k = 0;
        for(int k=1;k<n;k++)
        {
            if(nums2[k-1] > 1-k)ans++;
            mp[nums2[k-1]]--;
            ans -= mp[1-k];
            mp[nums2[n-1 + k]]++;
            if(ans > fin_ans)fin_ans=ans,fin_k=k;
        }
        return fin_k;
    }
};
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