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.

Approach #1: Array. [Java]

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

　　

Analysis:

The origainal order should be [0, 1, 2, 3, 4....], the number i should be on the position i. We just check the offset of each number, if the absolute value is large than 1, means the number of global inverion must be bigger than local inversion, because a local inversion is a global inversion, but a global one may not be local.

Proof:

If A[i] > i + 1, means at least one number that is smaller than A[i] is kicked out from first A[i] numbers, and the distance between this smaller number and A[i] is at least 2, then it is a non-local global inversion.

For example, A[i] = 3, i = 1, at least one number that is smaller than 3 is kicked out from first 3 numbers, and the distance between the smaller number and 3 is at least 2.

If A[i] < i - 1, means at least one number that is bigger than A[i] is kicked out from last n-i numbers, and the distance between this bigger number and A[i] is at least 2, then it is a non-local global inversion.

Reference:

https://leetcode.com/problems/global-and-local-inversions/discuss/113656/My-3-lines-C%2B%2B-Solution