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:

• 0 <= i < j < n
• nums[i] > nums[j]

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

• 0 <= i < n - 1
• nums[i] > nums[i + 1]

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:

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i] < n
• All the integers of nums are unique.
• nums is a permutation of all the numbers in the range [0, n - 1].

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;
    }
}

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