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_798.java
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_798.java
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package com.hit.basmath.learn.others;
/**
* 798. Smallest Rotation with Highest Score
* <p>
* Given an array A, we may rotate it by a non-negative integer K so that the array becomes A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1]. Afterward, any entries that are less than or equal to their index are worth 1 point.
* <p>
* For example, if we have [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].
* <p>
* Over all possible rotations, return the rotation index K that corresponds to the highest score we could receive. If there are multiple answers, return the smallest such index K.
* <p>
* Example 1:
* <p>
* Input: [2, 3, 1, 4, 0]
* Output: 3
* Explanation:
* Scores for each K are listed below:
* K = 0, A = [2,3,1,4,0], score 2
* K = 1, A = [3,1,4,0,2], score 3
* K = 2, A = [1,4,0,2,3], score 3
* K = 3, A = [4,0,2,3,1], score 4
* K = 4, A = [0,2,3,1,4], score 3
* So we should choose K = 3, which has the highest score.
* <p>
* Example 2:
* <p>
* Input: [1, 3, 0, 2, 4]
* Output: 0
* Explanation: A will always have 3 points no matter how it shifts.
* So we will choose the smallest K, which is 0.
* <p>
* Note:
* <p>
* A will have length at most 20000.
* A[i] will be in the range [0, A.length].
*/
public class _798 {
public int bestRotation(int[] A) {
int n = A.length;
int[] a = new int[n]; // to record interval start/end
for (int i = 0; i < A.length; i++) {
a[(i + 1) % n]++; // interval start
a[(i + 1 - A[i] + n) % n]--; // interval end
}
int count = 0;
int maxCount = -1;
int res = 0;
for (int i = 0; i < n; i++) { // find most overlap interval
count += a[i];
if (count > maxCount) {
res = i;
maxCount = count;
}
}
return res;
}
}