Last updated on October 9th, 2024 at 06:25 pm

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

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 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.

1. 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;
    }
};

2. 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;
    }
}

3. 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;
};

4. 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))