Solution Explanation for Find the Derangement of An Array

This problem asks to find the number of derangements of an array of integers from 1 to n, where a derangement is a permutation such that no element is in its original position. We'll explore two solutions: dynamic programming with and without space optimization.

Approach 1: Dynamic Programming

This approach uses dynamic programming to efficiently calculate the number of derangements.

State Definition:

Let dp[i] represent the number of derangements of an array of size i.

Base Cases:

• dp[0] = 1 (There's one way to derange an empty array)
• dp[1] = 0 (There's no way to derange an array of size 1)

Recurrence Relation:

For an array of size i, consider the placement of the element 1. We have i-1 choices for its position.

• Case 1: If we place 1 in position j, and then place j in position 1, we are left with i-2 elements to derange. This contributes (i-1) * dp[i-2] derangements.
• Case 2: If we place 1 in position j, but don't place j in position 1, we have i-1 elements left to derange. This contributes (i-1) * dp[i-1] derangements.

Therefore, the recurrence relation is:

dp[i] = (i - 1) * (dp[i - 1] + dp[i - 2])

Algorithm:

1. Initialize dp[0] and dp[1] with base cases.
2. Iterate from i = 2 to n, calculating dp[i] using the recurrence relation.
3. Return dp[n]. Remember to take the modulo with 10^9 + 7 to handle large results.

Time and Space Complexity:

• Time Complexity: O(n) - We iterate through the dp array once.
• Space Complexity: O(n) - We store the dp array.

Approach 2: Dynamic Programming with Space Optimization

We observe that the recurrence relation only depends on the previous two values (dp[i-1] and dp[i-2]). Therefore, we can optimize space by only storing these two values instead of the entire array.

Algorithm:

1. Initialize a = 1 and b = 0 (representing dp[0] and dp[1] respectively).
2. Iterate from i = 2 to n.
3. In each iteration, calculate the next value c = (i - 1) * (a + b) % mod.
4. Update a = b and b = c.
5. After the loop, b will hold the final result dp[n].

Time and Space Complexity:

• Time Complexity: O(n) - Same as before.
• Space Complexity: O(1) - We only store a constant number of variables.

Code Implementation (Python)

Approach 1:

class Solution:
    def findDerangement(self, n: int) -> int:
        mod = 10**9 + 7
        dp = [1, 0] + [0] * (n - 1)  # Initialize dp array
        for i in range(2, n + 1):
            dp[i] = (i - 1) * (dp[i - 1] + dp[i - 2]) % mod
        return dp[n]
 

Approach 2:

class Solution:
    def findDerangement(self, n: int) -> int:
        mod = 10**9 + 7
        a, b = 1, 0
        for i in range(2, n + 1):
            a, b = b, ((i - 1) * (a + b)) % mod
        return b

The code in other languages (Java, C++, Go) follows similar logic, adapting the syntax and data types accordingly as shown in the original response. The space-optimized version is generally preferred due to its lower memory usage, especially for large values of n.