Leetcode #2717: Semi-Ordered Permutation
In this guide, we solve Leetcode #2717 Semi-Ordered Permutation in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Array, Simulation
Intuition
The rules are explicit, so simulating the process step by step is safest.
Careful state updates prevent subtle bugs.
Approach
Translate the rules into state updates and apply them in order.
Track the final state or aggregate as required.
Steps:
- Translate rules into state updates.
- Iterate for each step.
- Return the final state.
Example
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.
Python Solution
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
Complexity
The time complexity is , where is the length of the array. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.