global and local inversions medium

Array/GlobalAndLocalInversions.py

@@ -0,0 +1,55 @@
+"""
+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
+