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.
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].
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.
Example 1:
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.
Example 2:
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.
Note:
A will have length at most 20000.
A[i] will be in the range [0, A.length].
class Solution(object):
def bestRotation(self, A):
"""
:type A: List[int]
:rtype: int
"""
cnts = collections.defaultdict(int)
ans = 0
for i, n in enumerate(A):
cnts[n - i] += 1
ans += n <= i
bestAns, bestIdx = ans, 0
for x in range(len(A) - 1, -1, -1):
y = len(A) - x - 1
if A[x] <= x + y: ans -= 1
if A[x] == 0: ans += 1
cnts[A[x] - x] -= 1
ans += cnts[y + 1]
if ans >= bestAns:
bestAns = ans
bestIdx = x
cnts[A[x] + y + 1] += 1
return
