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798. Smallest Rotation with Highest Score

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

  • 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;
        }
    }
    
  • 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;
        }
    };
    
  • 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
    
    
  • 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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