Algorithm
Problem Name: 798. Smallest Rotation with Highest Score
Problem Link: https://leetcode.com/problems/smallest-rotation-with-highest-score/
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 byk = 2, it becomes[1,3,0,2,4]. This is worth3points because1 > 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 <= 1050 <= nums[i] < nums.length
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int bestRotation(vector<int>& A) {
int n = A.size();
vector<int>k(n);
for(int i = 0; i < n; i++){
for(int j = 0; j < = i - A[i]; j++) k[j]++;
for(int j = i + 1; j < = i + n - A[i] && j < n; j++) k[j]++;
}
return max_element(k.begin(), k.end()) - k.begin();
}
};
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const bestRotation = function(A) {
const N = A.length
const bad = new Array(N).fill(0)
for (let i = 0; i < N; ++i) {
let left = (i - A[i] + 1 + N) % N
let right = (i + 1) % N
bad[left]--
bad[right]++
if (left > right) bad[0]--
}
let best = -N
let ans = 0,
cur = 0
for (let i = 0; i < N; ++i) {
cur += bad[i]
if (cur > best) {
best = cur
ans = i
}
}
return ans
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def bestRotation(self, A):
N = len(A)
change = [1] * N
for i in range(N): change[(i - A[i] + 1) % N] -= 1
for i in range(1, N): change[i] += change[i - 1]
return change.index(max(change))
Input