LeetCode 798: Smallest Rotation with Highest Score Solution

Master LeetCode problem 798 (Smallest Rotation with Highest Score), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

798. Smallest Rotation with Highest Score

Hard

Array Prefix Sum

Problem Explanation

Explanation

To solve this problem, we need to iterate through all possible rotations of the input array and calculate the score for each rotation. The score for a rotation is determined by counting how many elements are greater than their index after the rotation.

We can optimize the process by realizing that the score for each rotation depends only on the difference in indices between the current element and the element at index 0 after the rotation. We can use this insight to calculate the score for each rotation without actually rotating the array.

We can create an array diff to store the differences in indices between each element and the element at index 0 after rotation. Then, for each difference d in diff, we update the range [d, d + n - 1] to indicate that we have a point for this range.

Finally, we iterate through the diff array and calculate the score for each rotation. The rotation index k with the highest score is the answer.

The time complexity of this approach is O(n) where n is the length of the input array, and the space complexity is O(n).

Solution Code

class Solution {
    public int bestRotation(int[] nums) {
        int n = nums.length;
        int[] diff = new int[n];
        
        for (int i = 0; i < n; i++) {
            diff[(i - nums[i] + 1 + n) % n]--;
        }
        
        int maxScore = 0;
        int maxScoreIndex = 0;
        int score = 0;
        
        for (int i = 0; i < n; i++) {
            score += diff[i];
            if (score > maxScore) {
                maxScore = score;
                maxScoreIndex = i;
            }
        }
        
        return maxScoreIndex;
    }
}

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