source: https://leetcode.com/problems/smallest-rotation-with-highest-score/
C/C++ Solution to LeetCode problem 798. Smallest Rotation with Highest Score.
Problem
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
.
Examples
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
Solution
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class Solution {
public:
int bestRotation(vector<int>& nums) {
vector<int> points(2 * nums.size() + 1, 0);
int k = 0;
int pMax = 0;
for (int i = 0; i < nums.size(); i += 1) {
if (nums[i] <= i) {
points[0] += 1;
points[(i - nums[i]) + 1] -= 1;
}
points[i + 1] += 1;
points[i + (nums.size() - nums[i]) + 1] -= 1;
}
pMax = points[0];
for (int i = 1; i < points.size(); i += 1) {
points[i] += points[i - 1];
if (points[i] > pMax) {
pMax = points[i];
k = i;
}
}
return k;
}
};