<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:
#define pi pair<int,int>

    int bestRotation(vector<int>& nums) {
        int n=nums.size();
        priority_queue<pi,vector<pi>,greater<pi>> pq;
        for(int i=0;i<n;i++){
            int start=0,end=0;
            if(nums[i]>i){
                start=i+1;
                end=min(n-1,n-nums[i]+i);
            }
            else if(nums[i]<=i){
                int start1=0,end1=0;
                end1=i-nums[i];
                pq.push({start1,end1});
                // cout<<i<<" "<<start1<<" "<<end1<<endl;
                if(end1==n-1) continue;
                start=i+1;
                end=n-1;
            }
            // cout<<i<<" "<<start<<" "<<end<<endl;
            pq.push({start,end});
        }
        int ans=0,maxi=0,num=0;
        unordered_map<int,int> map;
        for(int i=0;i<n;i++){
            if(!pq.size()) break;
            // int count=0;
            while(pq.size() && pq.top().first==i){
                map[pq.top().second]++;
                num++;
                // if(pq.top().second==i) count++;
                pq.pop();
            }
            if(maxi<num){
                num=maxi;
                ans=i;
            }
            num-=map[i];
        }
        return ans;
    }
};
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