LeetCode - Global and Local Inversions

Question Definition

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.

  Java Solution

  public boolean isIdealPermutation(int[] A) {
    if (A.length < 2) return true;
    for (int i = 1; i < A.length; i++) {
        if (A[i-1] > A[i])
        {
            int temp = A[i];
            A[i] = A[i-1];
            A[i-1] = temp;
            if (A[i-1] != i-1 || A[i] != i) return false;
        }
    }
    return true;
  }