2717. Semi-Ordered Permutation 

Description

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] <= 50
nums is a permutation.

Solutions

Solution 1: Find the Positions of 1 and n

We can first find the indices $i$ and $j$ of $1$ and $n$, respectively. Then, based on the relative positions of $i$ and $j$, we can determine the number of swaps required.

If $i < j$, the number of swaps required is $i + n - j - 1$. If $i > j$, the number of swaps required is $i + n - j - 2$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

class Solution:
    def semiOrderedPermutation(self, nums: List[int]) -> int:
        n = len(nums)
        i = nums.index(1)
        j = nums.index(n)
        k = 1 if i < j else 2
        return i + n - j - k

Java

class Solution {
    public int semiOrderedPermutation(int[] nums) {
        int n = nums.length;
        int i = 0, j = 0;
        for (int k = 0; k < n; ++k) {
            if (nums[k] == 1) {
                i = k;
            }
            if (nums[k] == n) {
                j = k;
            }
        }
        int k = i < j ? 1 : 2;
        return i + n - j - k;
    }
}

C++

class Solution {
public:
    int semiOrderedPermutation(vector<int>& nums) {
        int n = nums.size();
        int i = find(nums.begin(), nums.end(), 1) - nums.begin();
        int j = find(nums.begin(), nums.end(), n) - nums.begin();
        int k = i < j ? 1 : 2;
        return i + n - j - k;
    }
};

Go

func semiOrderedPermutation(nums []int) int {
	n := len(nums)
	var i, j int
	for k, x := range nums {
		if x == 1 {
			i = k
		}
		if x == n {
			j = k
		}
	}
	k := 1
	if i > j {
		k = 2
	}
	return i + n - j - k
}

TypeScript

function semiOrderedPermutation(nums: number[]): number {
    const n = nums.length;
    const i = nums.indexOf(1);
    const j = nums.indexOf(n);
    const k = i < j ? 1 : 2;
    return i + n - j - k;
}

Rust

impl Solution {
    pub fn semi_ordered_permutation(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let (mut i, mut j) = (0, 0);

        for k in 0..n {
            if nums[k] == 1 {
                i = k;
            }
            if nums[k] == (n as i32) {
                j = k;
            }
        }

        let k = if i < j { 1 } else { 2 };
        (i + n - j - k) as i32
    }
}