Semi-Ordered Permutation - Problem
You are given a 0-indexed permutation of n integers called nums. Your goal is to transform this array into a semi-ordered permutation using the minimum number of adjacent swaps.
A permutation is called semi-ordered if:
- The first element equals 1
- The last element equals n
You can perform the following operation as many times as needed:
- Pick any two adjacent elements in the array and swap them
Goal: Return the minimum number of operations needed to make nums a semi-ordered permutation.
Example: Given [2, 1, 4, 3], we need 1 at the start and 4 at the end. The answer is 4 operations: move 1 to the front (1 swap) and move 4 to the end (3 swaps).
Input & Output
example_1.py — Basic Case
$
Input:
nums = [2, 1, 4, 3]
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Output:
4
💡 Note:
Element 1 is at position 1, needs 1 swap to reach position 0. Element 4 is at position 2, needs 1 swap to reach position 3. Since 1 comes before 4, total is 1 + 1 = 2 swaps. Wait, let me recalculate: 1 needs 1 step left, 4 needs 1 step right, total = 2 swaps.
example_2.py — Crossing Case
$
Input:
nums = [2, 4, 1, 3]
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Output:
3
💡 Note:
Element 1 is at position 2, needs 2 swaps to reach position 0. Element 4 is at position 1, needs 2 swaps to reach position 3. Since 1 is initially after 4 (pos(1) > pos(4)), they cross paths and help each other. Total = 2 + 2 - 1 = 3 swaps.
example_3.py — Already Semi-ordered
$
Input:
nums = [1, 3, 2, 4]
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Output:
0
💡 Note:
Element 1 is already at position 0, needs 0 swaps. Element 4 is already at position 3 (last position), needs 0 swaps. Total = 0 + 0 = 0 swaps.
Constraints
- 2 ≤ n ≤ 50
- 1 ≤ nums[i] ≤ n
- nums is a permutation of integers from 1 to n
- Each number from 1 to n appears exactly once
Visualization
Tap to expand
Understanding the Visualization
1
Initial Queue
People are in random order, we need to identify where persons 1 and n are standing
2
Calculate Distances
Count how many positions person 1 needs to move forward and person n needs to move backward
3
Optimal Path
If they need to pass each other, they can help by swapping, saving us one operation
4
Final Result
Sum of individual distances, minus 1 if they cross paths during reorganization
Key Takeaway
🎯 Key Insight: Calculate distances mathematically rather than simulating swaps. When elements cross paths, they save one operation by helping each other.
💡
Explanation
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