# Problem

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.

# Solution1: Brute Force (TLE)

Time complexity: O(n^2)

Space complexity: O(1)

C++

# Solution2: MergeSort

Time complexity: O(nlogn)

Space complexity: O(n)

C#

# Solution3: Input Property

Input is a permutation of [0, 1, …, N – 1]

Time Complexity: O(n)

Space Complexity: O(1)

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