2717. Semi-Ordered Permutation

Problem

You are given a 0-indexed permutation of n integers nums.

A permutation is called semi-ordered if the first number equals 1 and the last number equals n. You can perform the below operation as many times as you want until you make nums a semi-ordered permutation:

• Pick two adjacent elements in nums, then swap them.

Return **the minimum number of operations to make *nums* a **semi-ordered permutation****.

A permutation is a sequence of integers from 1 to n of length n containing each number exactly once.

Example 1:

Input: nums = [2,1,4,3]
Output: 2
Explanation: We can make the permutation semi-ordered using these sequence of operations:
1 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].
2 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].
It can be proved that there is no sequence of less than two operations that make nums a semi-ordered permutation.

Example 2:

Input: nums = [2,4,1,3]
Output: 3
Explanation: We can make the permutation semi-ordered using these sequence of operations:
1 - swap i = 1 and j = 2. The permutation becomes [2,1,4,3].
2 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].
3 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].
It can be proved that there is no sequence of less than three operations that make nums a semi-ordered permutation.

Example 3:

Input: nums = [1,3,4,2,5]
Output: 0
Explanation: The permutation is already a semi-ordered permutation.

Constraints:

• 2 <= nums.length == n <= 50
• 1 <= nums[i]&nbsp;<= 50
• nums is a permutation.

Solution (Java)

class Solution {
    public int semiOrderedPermutation(int[] A) {
        int n = A.length, i = 0, j = 0;
        for (int k = 0; k < n; k++) {
            if (A[k] == 1) i = k;
            if (A[k] == A.length) j = k;
        }
        return i + n - 1 - j - (i > j ? 1 : 0);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).