GlobalAndLocalInversions.py

"""
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 1:

Input: A = [1,0,2]
Output: true
Explanation: There is 1 global inversion, and 1 local inversion.
Example 2:

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.

思路：
 二分法。

测试地址：
https://leetcode.com/contest/weekly-contest-69/problems/global-and-local-inversions/

"""
import bisect

class Solution(object):
    def isIdealPermutation(self, A):
        """
        :type A: List[int]
        :rtype: bool
        """
        global_inversions = []
        _g = 0
        
        for i in A[::-1]:
            _g += bisect.bisect_left(global_inversions, i)
            bisect.insort_left(global_inversions, i)
        
        _l = 0
        
        for i in range(len(A)-1):
            if A[i] > A[i+1]:
                _l += 1

        return _g == _l