LeetCode 0798. Smallest Rotation with Highest Score Solution in Java, C++, Python & Go | Explanation + Code

Description

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[0], nums[1], ..., 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.

 

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

Solutions

Solution 1

PythonJavaC++Go

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(code-box)

class Solution { public int bestRotation(int[] nums) { int n = nums.length; int[] d = new int[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 mx = -1; int s = 0; int ans = n; for (int k = 0; k < n; ++k) { s += d[k]; if (s > mx) { mx = s; ans = k; } } return ans; } }(code-box)

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; } };(code-box)

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 }(code-box)