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.

Solution

Java Program

class Solution {
    public boolean isIdealPermutation(int[] A) {
        
        //O(N) Time Complexity
        
        int currentMax = 0;
        for(int index = 0; index < A.length-2; index++)
        {
            //Check currentMax greater than A[index+2]
            //If Greater, then it contribute to Global versin
            //But doesn't contribute to Local Version

            currentMax = Math.max(currentMax,A[index]);
            
            if(currentMax > A[index+2])
                return false;
        }
        return true;
    }
}

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