Find the Derangement of An Array - Problem
Derangement Problem: The Perfect Shuffle Challenge
In combinatorial mathematics, a derangement is a fascinating permutation where every element is displaced from its original position. Think of it as the "perfect shuffle" where nothing ends up where it started!
🎯 Your Mission: Given an integer
📊 Example: For
• Valid derangement:
• Invalid:
⚠️ Important: Since the result can be astronomically large, return the answer modulo
In combinatorial mathematics, a derangement is a fascinating permutation where every element is displaced from its original position. Think of it as the "perfect shuffle" where nothing ends up where it started!
🎯 Your Mission: Given an integer
n, calculate how many derangements are possible for an array containing integers from 1 to n in ascending order.📊 Example: For
n = 3 with array [1, 2, 3]:• Valid derangement:
[2, 3, 1] ✅ (no element at original position)• Invalid:
[1, 3, 2] ❌ (element 1 is still at position 0)⚠️ Important: Since the result can be astronomically large, return the answer modulo
109 + 7. Input & Output
example_1.py — Basic Case
$
Input:
n = 3
›
Output:
2
💡 Note:
For array [1,2,3], the valid derangements are [2,3,1] and [3,1,2]. In both cases, no element appears at its original position.
example_2.py — Edge Case
$
Input:
n = 1
›
Output:
0
💡 Note:
With only one element [1], it's impossible to create a derangement since the element must stay in its original position.
example_3.py — Larger Case
$
Input:
n = 4
›
Output:
9
💡 Note:
For array [1,2,3,4], there are 9 valid derangements out of 24 total permutations. The derangements ensure no element i is at position i.
Visualization
Tap to expand
Understanding the Visualization
1
Setup
n guests check their hats numbered 1 to n
2
Constraint
No guest can receive their own hat back
3
Count
How many ways can we return all hats incorrectly?
4
Formula
D(n) = (n-1) × [D(n-1) + D(n-2)]
Key Takeaway
🎯 Key Insight: The derangement formula elegantly captures the combinatorial structure: choose any of (n-1) positions for the first element, then either create a swap or recursively solve smaller subproblems.
Time & Space Complexity
Time Complexity
O(n)
Single pass through n iterations
✓ Linear Growth
Space Complexity
O(1)
Only storing previous two values
✓ Linear Space
Constraints
- 1 ≤ n ≤ 106
- Return result modulo 109 + 7
- Time limit: 1 second
💡
Explanation
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