# 798. Smallest Rotation with Highest Score

Given an array `A`, we may rotate it by a non-negative integer `K` so that the array becomes `A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1]`.  Afterward, any entries that are less than or equal to their index are worth 1 point.

For example, if we have `[2, 4, 1, 3, 0]`, and we rotate by `K = 2`, it becomes `[1, 3, 0, 2, 4]`.  This is worth 3 points because 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point].

Over all possible rotations, return the rotation index K that corresponds to the highest score we could receive.  If there are multiple answers, return the smallest such index K.

``````Example 1:
Input: [2, 3, 1, 4, 0]
Output: 3
Explanation:
Scores for each K are listed below:
K = 0,  A = [2,3,1,4,0],    score 2
K = 1,  A = [3,1,4,0,2],    score 3
K = 2,  A = [1,4,0,2,3],    score 3
K = 3,  A = [4,0,2,3,1],    score 4
K = 4,  A = [0,2,3,1,4],    score 3
``````

So we should choose K = 3, which has the highest score.

``````Example 2:
Input: [1, 3, 0, 2, 4]
Output: 0
Explanation:  A will always have 3 points no matter how it shifts.
So we will choose the smallest K, which is 0.
``````

Note:

• `A` will have length at most `20000`.
• `A[i]` will be in the range `[0, A.length]`.

A:    2   3   1   4   0   (K = 0)

A:    3   1   4   0   2   (K = 1)

A:    1   4   0   2   3   (K = 2)

A:    4   0   2   3   1   (K = 3)

A:    0   2   3   1   4   (K = 4)

idx:  0   1   2   3   4

``````class Solution {
public:
int bestRotation(vector<int>& A) {
int n = A.size();
vector<int> change(n, 0);
for (int i = 0; i < n; ++i) change[(i - A[i] + 1 + n) % n] -= 1;
for (int i = 1; i < n; ++i) change[i] += change[i - 1] + 1;
return max_element(change.begin(), change.end()) - change.begin();
}
};
``````

``````class Solution {
public:
int bestRotation(vector<int>& A) {
int n = A.size(), res = 0;
vector<int> change(n, 0);
for (int i = 0; i < n; ++i) change[(i - A[i] + 1 + n) % n] -= 1;
for (int i = 1; i < n; ++i) {
change[i] += change[i - 1] + 1;
res = (change[i] > change[res]) ? i : res;
}
return res;
}
};
``````

https://leetcode.com/problems/smallest-rotation-with-highest-score/discuss/121296/7-lines-C++-solution

https://leetcode.com/problems/smallest-rotation-with-highest-score/discuss/118725/Easy-and-Concise-5-lines-Solution-C++JavaPython?page=2

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