Here, We see Smallest Rotation with Highest Score LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
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Smallest Rotation with Highest Score LeetCode Solution
Table of Contents
Problem Statement
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 worth3
points 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.
Smallest Rotation with Highest Score Leetcode Solution C++
class Solution {
public:
int bestRotation(vector<int>& nums) {
const int N = nums.size();
vector<int> tmp(N+1, 0);
for (int i(0); i < N; ++i) {
if (nums[i] <= i) {
tmp[0] += 1;
tmp[i-nums[i]+1] -= 1;
tmp[i+1] += 1;
tmp[N] -= 1;
}
else {
tmp[i+1] += 1;
tmp[N-nums[i]+i+1] -= 1;
}
}
int maxc = INT_MIN;
int maxK = -1;
int cur = 0;
for (int i(0); i < N; ++i) {
cur += tmp[i];
if (cur > maxc) {
maxc = cur;
maxK = i;
}
}
return maxK;
}
};
Code language: PHP (php)
Smallest Rotation with Highest Score Leetcode Solution Java
class Solution {
public int bestRotation(int[] nums) {
final int size = nums.length;
int[] rsc = new int[size];
for(int i = 0; i < size - 1; i++) {
int value = nums[i];
int downPos = (i + 1 + size - value) % size;
rsc[downPos]--;
}
int value = nums[size-1];
if( value != 0 ) rsc[size - value]--;
int bestk = 0;
int bestscore = rsc[0];
int score = rsc[0];
for(int i = 1; i < nums.length; i++) {
score += rsc[i] + 1;
if( score > bestscore ) {
bestk = i;
bestscore = score;
}
}
return bestk;
}
}
Code language: PHP (php)
Smallest Rotation with Highest Score Leetcode Solution JavaScript
var bestRotation = function (nums) {
let acc = new Uint32Array(100001);
let n = nums.length;
acc.fill(0, 0, n + 1);
for (let i = 0; i < n; ++i) {
++acc[Math.min(i + 1, Math.max(0, i - nums[i] + 1))];
}
let bestInd = 0;
let bestCnt = n - acc[0];
let currCnt = bestCnt;
for (let i = 1; i < n; ++i) {
currCnt -= acc[i];
if (nums[i - 1] < n) {
++currCnt;
let exp = i + n - nums[i - 1];
if (exp < n) ++acc[exp];
}
if (currCnt > bestCnt) {
bestCnt = currCnt;
bestInd = i;
}
}
return bestInd;
};
Code language: JavaScript (javascript)
Smallest Rotation with Highest Score Leetcode Solution Python
class Solution(object):
def bestRotation(self, nums):
n = len(nums)
change = [1] * n
for i, num in enumerate(nums):
change[(i - num + 1) % n] -= 1
for i in range(1, n):
change[i] += change[i - 1]
return change.index(max(change))