LeetCode in Kotlin
2717. Semi-Ordered Permutation
Easy
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 <= 501 <= nums[i] <= 50nums is a permutation.
Solution
class Solution {
fun semiOrderedPermutation(nums: IntArray): Int {
val a = 1
var b = nums.size
var idxA = 0
var idxB = 0
for (i in nums.indices) {
if (nums[i] == a) idxA = i
if (nums[i] == b) idxB = i
}
b = b - idxB - 1
return if (idxB < idxA) idxA + b - 1 else idxA + b
}
}