775. 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.

class Solution {
    public boolean isIdealPermutation(int[] A) {
        int glb = 0, loc = 0;
        for(int i = 0; i < A.length - 1; i++) {
            if(A[i] > A[i + 1]) loc++;
            for(int j = i + 1; j < A.length; j++) {
                if(A[i] > A[j]) glb++;
            }
        }
        return glb == loc;
    }
}

O(n2), TLE

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;
    }
}

 

local是前大于后，比如2>1, 但是如果是3,2,1，这时候local是2，global已经是3了，所以不行。1,0,2这种才行。所以有如下：

 是local一定是global，因为global很容易，所以要返回true一定不能找到max(A[i]) > A[i + 2], 持续找，如果找不到就返回true，否则返回false

 

 

public boolean isIdealPermutation(int[] A) {

        for (int i = 0; i < A.length; i++) {
            if (Math.abs(i - A[i]) > 1)
                return false;
        }

        return true;
    }

 https://leetcode.com/problems/global-and-local-inversions/discuss/1143422/JS-Python-Java-C%2B%2B-or-Simple-3-Line-Solutions-w-Explanation