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Global and Local Inversions

Updated: Apr 6, 2021

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

Solution:


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

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