How do you solve Semi-Ordered Permutation on LeetCode?

Semi-Ordered Permutation (LeetCode #2717) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The positions of 1 and n are crucial in determining the number of swaps needed.

What is the brute force solution for Semi-Ordered Permutation?

The brute force approach for Semi-Ordered Permutation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the process of swapping adjacent elements until the permutation meets the semi-ordered condition. This is straightforward but inefficient, as it may require many swaps. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Semi-Ordered Permutation?

Semi-Ordered Permutation uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Semi-Ordered Permutation?

Always think about edge cases, such as when the array is already semi-ordered. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice identifying patterns in problems, as many can be solved with similar strategies.

What are common mistakes when solving Semi-Ordered Permutation?

Not considering the case where 1 and n are already in the correct positions. Overcomplicating the swapping process instead of calculating based on indices.

#2717

Semi-Ordered Permutation

Easy

ArraySimulationArraySimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the positions of 1 and n to calculate the minimum swaps needed directly, avoiding unnecessary iterations. This is efficient and straightforward.

⚙️

Algorithm

3 steps

1Step 1: Find the index of 1 (let's call it x) and the index of n (let's call it y).
2Step 2: If x < y, the answer is x + (n - y - 1).
3Step 3: If x > y, the answer is x + (n - y - 1) - 1.

solution.py11 lines

1# Full working Python code
2
3def min_swaps(nums):
4    n = len(nums)
5    x = nums.index(1)
6    y = nums.index(n)
7    if x < y:
8        return x + (n - y - 1)
9    else:
10        return x + (n - y - 1) - 1
11

ℹ

Complexity note: This solution is linear because we only traverse the array a couple of times to find the indices of 1 and n, making it very efficient.

1The positions of 1 and n are crucial in determining the number of swaps needed.
2The order of elements between 1 and n does not affect the swap count as long as they are in the correct positions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.