g0701_0800.s0775_global_and_local_inversions.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0701_0800.s0775_global_and_local_inversions;

// #Medium #Array #Math #2022_03_26_Time_1_ms_(99.74%)_Space_51.6_MB_(90.31%)

/**
 * 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]`.
**/
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
     * 
     * 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;
    }
}