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Formatted question description: https://leetcode.ca/all/798.html

798. Smallest Rotation with Highest Score

Level

Hard

Description

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].

Solution

For index i, if the rotation index K is between i - A[i] + 1 and i (both are mod A.length), then A[i] will have no points. Create an array points of length A.length. For each index i, subtract 1 from points[i - A[i] + 1] and add 1 to points[i + 1]. Then loop over points and calculate the sums of each prefix. If a prefix sum is the greatest, then update the maximum sum and the corresponding index. Finally, return the index.

  • class Solution {
    public int bestRotation(int[] A) {
        int length = A.length;
        int[] points = new int[length];
        for (int i = 0; i < length; i++) {
            int low = (i - A[i] + 1 + length) % length;
            int high = (i + 1) % length;
            points[low]--;
            points[high]++;
            if (low > high)
                points[0]--;
        }
        int maxIndex = 0;
        int maxScore = -length;
        int curScore = 0;
        for (int i = 0; i < length; i++) {
            curScore += points[i];
            if (curScore > maxScore) {
                maxIndex = i;
                maxScore = curScore;
            }
        }
        return maxIndex;
    }
}

  • class Solution:
    def bestRotation(self, nums: List[int]) -> int:
        n = len(nums)
        mx, ans = -1, n
        d = [0] * n
        for i, v in enumerate(nums):
            l, r = (i + 1) % n, (n + i + 1 - v) % n
            d[l] += 1
            d[r] -= 1
        s = 0
        for k, t in enumerate(d):
            s += t
            if s > mx:
                mx = s
                ans = k
        return ans



  • class Solution {
public:
    int bestRotation(vector<int>& nums) {
        int n = nums.size();
        int mx = -1, ans = n;
        vector<int> d(n);
        for (int i = 0; i < n; ++i) {
            int l = (i + 1) % n;
            int r = (n + i + 1 - nums[i]) % n;
            ++d[l];
            --d[r];
        }
        int s = 0;
        for (int k = 0; k < n; ++k) {
            s += d[k];
            if (s > mx) {
                mx = s;
                ans = k;
            }
        }
        return ans;
    }
};

  • func bestRotation(nums []int) int {
	n := len(nums)
	d := make([]int, n)
	for i, v := range nums {
		l, r := (i+1)%n, (n+i+1-v)%n
		d[l]++
		d[r]--
	}
	mx, ans, s := -1, n, 0
	for k, t := range d {
		s += t
		if s > mx {
			mx = s
			ans = k
		}
	}
	return ans
}

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