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''' | ||
We have some permutation A of [0, 1, ..., N - 1], where | ||
N is the length of A. | ||
The number of (global) inversions is the number of i < j | ||
with 0 <= i < j < N and A[i] > A[j]. | ||
The number of local inversions is the number of i with | ||
0 <= i < N and A[i] > A[i+1]. | ||
Return true if and only if the number of global inversions | ||
is equal to the number of local inversions. | ||
Example: | ||
Input: A = [1,0,2] | ||
Output: true | ||
Explanation: There is 1 global inversion, and 1 local | ||
inversion. | ||
Example: | ||
Input: A = [1,2,0] | ||
Output: false | ||
Explanation: There are 2 global inversions, and 1 local | ||
inversion. | ||
Note: | ||
- A will be a permutation of [0, 1, ..., A.length - 1]. | ||
- A will have length in range [1, 5000]. | ||
- The time limit for this problem has been reduced. | ||
''' | ||
#Difficulty: Medium | ||
# | ||
|
||
#Runtime: 324 ms, faster than 86.59% of Python3 online submissions for Global and Local Inversions. | ||
#Memory Usage: 15.1 MB, less than 44.06% of Python3 online submissions for Global and Local Inversions. | ||
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||
class Solution: | ||
def isIdealPermutation(self, A: List[int]) -> bool: | ||
for i in range(len(A)): | ||
if abs(A[i] - i) > 1: | ||
return False | ||
return True |
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