Smallest Rotation with Highest Score
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# LeetCode #798: Smallest Rotation with Highest Score (C/C++).

hard

source: https://leetcode.com/problems/smallest-rotation-with-highest-score/
C/C++ Solution to LeetCode problem 798. Smallest Rotation with Highest Score.

## Problem

You are given an array nums. You can rotate it by a non-negative integer k so that the array becomes [nums[k], nums[k + 1], ... nums[nums.length - 1], nums, nums, ..., nums[k-1]]. Afterward, any entries that are less than or equal to their index are worth one point.

• For example, if we have nums = [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].

Return the rotation index k that corresponds to the highest score we can achieve if we rotated nums by it. If there are multiple answers, return the smallest such index k.

## Examples

### Example 1:

Input: nums = [2,3,1,4,0]
Output: 3
Explanation: 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.

### Example 2:

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

## Constraints

• 1 <= nums.length <= 105
• 0 <= nums[i] < nums.length

## Solution

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 class Solution { public: int bestRotation(vector<int>& nums) { vector<int> points(2 * nums.size() + 1, 0); int k = 0; int pMax = 0; for (int i = 0; i < nums.size(); i += 1) { if (nums[i] <= i) { points += 1; points[(i - nums[i]) + 1] -= 1; } points[i + 1] += 1; points[i + (nums.size() - nums[i]) + 1] -= 1; } pMax = points; for (int i = 1; i < points.size(); i += 1) { points[i] += points[i - 1]; if (points[i] > pMax) { pMax = points[i]; k = i; } } return k; } };