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].

# @lc code=start
using LeetCode

function best_rotation(nums::Vector{Int})::Int
    n = length(nums)
    diffs = fill(0, n)
    for (i, num) in enumerate(nums)
        diffs[mod(i - num - 1, n) + 1] -= 1
        diffs[i] += 1
    end
    ans = maxscore = score = 0
    for (i, num) in enumerate(@view(diffs[1:(end - 1)]))
        score += num
        if score > maxscore
            maxscore, ans = score, i
        end
    end
    return ans
end
# @lc code=end

best_rotation (generic function with 1 method)

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