LeetCode-in-Java

775. Global and Local Inversions

Medium

You are given an integer array nums of length n which represents a permutation of all the integers in the range [0, n - 1].

The number of global inversions is the number of the different pairs (i, j) where:

The number of local inversions is the number of indices i where:

Return true if the number of global inversions is equal to the number of local inversions.

Example 1:

Input: nums = [1,0,2]

Output: true

Explanation: There is 1 global inversion and 1 local inversion.

Example 2:

Input: nums = [1,2,0]

Output: false

Explanation: There are 2 global inversions and 1 local inversion.

Constraints:

Solution

public class Solution {
    /*
     * from the above solution, we can tell that if we can find the minimum of A[j] where j >= i + 2,
     * then we could quickly return false, so two steps:
     * 1. remembering minimum
     * 2. scanning from right to left
     * <p>
     * Time: O(n)
     */
    public boolean isIdealPermutation(int[] nums) {
        int min = nums.length;
        for (int i = nums.length - 1; i >= 2; i--) {
            min = Math.min(min, nums[i]);
            if (nums[i - 2] > min) {
                return false;
            }
        }
        return true;
    }
}