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634. Find the Derangement of An Array

Approach 1: Top-down

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int findDerangement(int n) {
    vector<int> mem(n + 1);
    return findDerangement(n, mem);
  }

 private:
  static constexpr int kMod = 1'000'000'007;

  int findDerangement(int i, vector<int>& mem) {
    if (i == 0)
      return 1;
    if (i == 1)
      return 0;
    if (mem[i] > 0)
      return mem[i];
    return mem[i] = (i - 1L) *
                    (findDerangement(i - 1, mem) +  //
                     findDerangement(i - 2, mem)) %
                    kMod;
  }
};

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class Solution {
  public int findDerangement(int n) {
    int[] mem = new int[n + 1];
    return findDerangement(n, mem);
  }

  private static final int kMod = 1_000_000_007;

  private int findDerangement(int i, int[] mem) {
    if (i == 0)
      return 1;
    if (i == 1)
      return 0;
    if (mem[i] > 0)
      return mem[i];
    return mem[i] = (int) ((i - 1L) *
                           (findDerangement(i - 1, mem) + //
                            findDerangement(i - 2, mem)) %
                           kMod);
  }
}

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class Solution:
  def findDerangement(self, n: int) -> int:
    kMod = 1_000_000_007

    @functools.lru_cache(None)
    def dp(i: int) -> int:
      if i == 0:
        return 1
      if i == 1:
        return 0
      return (i - 1) * (dp(i - 1) + dp(i - 2)) % kMod

    return dp(n)

Approach 2: Bottom-up

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int findDerangement(int n) {
    constexpr int kMod = 1'000'000'007;
    vector<int> dp(n + 1);

    dp[0] = 1;

    for (int i = 2; i <= n; ++i)
      dp[i] = (i - 1L) * (dp[i - 1] + dp[i - 2]) % kMod;

    return dp[n];
  }
};

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class Solution {
  public int findDerangement(int n) {
    final int kMod = 1_000_000_007;
    int[] dp = new int[n + 1];

    dp[0] = 1;

    for (int i = 2; i <= n; ++i)
      dp[i] = (int) ((i - 1L) * (dp[i - 1] + dp[i - 2]) % kMod);

    return dp[n];
  }
}

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class Solution:
  def findDerangement(self, n: int) -> int:
    kMod = 1_000_000_007
    dp = [1] + [0] * n

    for i in range(2, n + 1):
      dp[i] = (i - 1) * (dp[i - 1] + dp[i - 2]) % kMod

    return dp[n]