Global and Local Inversions

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.

// The local and global adjacent inversions are same
// therefore we are only looking for an inversion at min distance of 2
class Solution {
    public boolean isIdealPermutation(int[] A) {
        int cmax = 0;
        for (int i = 0; i < A.length - 2; ++i) {
            cmax = Math.max(cmax, A[i]);
            if (cmax > A[i + 2])
                return false;
        }
        return true;
    }
}