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.

思路：这个题目brute force很好想，就是count，再比较。但是注意到题目只要返回true or false，所以只要遇见不满足的情况就可以直接返回false了。那么什么情况下会返回false呢？

local inversion <= global inversion, 什么情况下 local < global 呢？就是 i < j && a[i] > a[j] ，只要除掉i, i+1比较。

那么换句话说，对于每个i，就可以判断 [0 ~ i-2] 有没有大于a[i] 的即可，如果有，则有inversion, time complexity: O(N)

class Solution {
    public boolean isIdealPermutation(int[] A) {
        if(A == null || A.length == 0) return true;
        int max = -1;
        for(int i=0; i<A.length-2; i++){
            max = Math.max(max, A[i]);
            if(max > A[i+2]){
                return false;
            }
        }
        return true;
    }
}