<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:
    static const int maxN = 1e5;
    int seg[4*maxN + 5];
    int lazy[4*maxN + 5];
    void update(int l,int r,int uL,int uR,int ind,int x){
        if(lazy[ind] != 0){
            seg[ind] += lazy[ind];
            if(l != r){
                lazy[2*ind+1] += lazy[ind];
                lazy[2*ind+2] += lazy[ind];
            }
            lazy[ind] = 0;
            
        }
        if(uL>r || uR < l)return;
        if(uL <= l && uR >= r){
            seg[ind] += x;
            if(l != r){
                lazy[2*ind+1] += x;
                lazy[2*ind+2] += x;
            }
            return;
        }
        int mid = l + (r-l)/2;
        update(l,mid,uL,uR,2*ind+1,x);
        update(mid+1,r,uL,uR,2*ind+2,x);
        seg[ind]  = max(seg[2*ind+1],seg[2*ind+2]); 
    }
    int query(int l,int r,int uL,int uR,int ind){
        if(lazy[ind] != 0){
            seg[ind] += lazy[ind];
            if(l != r){
                lazy[2*ind+1] += lazy[ind];
                lazy[2*ind+2] += lazy[ind];
            }
            lazy[ind] = 0;
        }
        if(uL>r || uR < l)return INT_MIN;
        if(uL <= l && uR >= r)return seg[ind];
        int mid = l + (r-l)/2;
        int left = query(l,mid,uL,uR,2*ind+1);
        int right = query(mid+1,r,uL,uR,2*ind+2);
        return max(left,right);
    }
    
    int bestRotation(vector<int>& nums) {
        int n = nums.size();
        memset(seg,0,sizeof(seg[0])*n);
        memset(lazy,0,sizeof(lazy[0])*n);
        for(int i = 0;i<nums.size();i++){
            int leftShiftAdd =  i - nums[i];
            if(leftShiftAdd>=0){

                update(0,n-1,0,leftShiftAdd,0,1);
            }
            if(i == n-1)continue;
            int rightShiftLow = max(i+1,nums[i]);
            int reachUp =  n - (rightShiftLow - i);
            int reachDown =  n - (n-1 - i);

            update(0,n-1,reachDown,reachUp,0,1);
			
        }

        int maxi = 0;
        int ind = -1;
        for(int i = 0;i<n;i++){
            int a = query(0,n-1,i,i,0);
            if(maxi< a){
                maxi = a;
                ind = i;
            }
        }

        
        return ind;
    }
};
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