Description#
In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position.
You are given an integer n
. There is originally an array consisting of n
integers from 1
to n
in ascending order, return the number of derangements it can generate. Since the answer may be huge, return it modulo 109 + 7
.
Example 1:
Input: n = 3
Output: 2
Explanation: The original array is [1,2,3]. The two derangements are [2,3,1] and [3,1,2].
Example 2:
Input: n = 2
Output: 1
Constraints:
Solutions#
Solution 1#
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| class Solution:
def findDerangement(self, n: int) -> int:
mod = 10**9 + 7
f = [1] + [0] * n
for i in range(2, n + 1):
f[i] = (i - 1) * (f[i - 1] + f[i - 2]) % mod
return f[n]
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| class Solution {
public int findDerangement(int n) {
long[] f = new long[n + 1];
f[0] = 1;
final int mod = (int) 1e9 + 7;
for (int i = 2; i <= n; ++i) {
f[i] = (i - 1) * (f[i - 1] + f[i - 2]) % mod;
}
return (int) f[n];
}
}
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| class Solution {
public:
int findDerangement(int n) {
long long f[n + 1];
memset(f, 0, sizeof(f));
f[0] = 1;
const int mod = 1e9 + 7;
for (int i = 2; i <= n; i++) {
f[i] = (i - 1LL) * (f[i - 1] + f[i - 2]) % mod;
}
return f[n];
}
};
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| func findDerangement(n int) int {
f := make([]int, n+1)
f[0] = 1
const mod = 1e9 + 7
for i := 2; i <= n; i++ {
f[i] = (i - 1) * (f[i-1] + f[i-2]) % mod
}
return f[n]
}
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Solution 2#
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| 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
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| class Solution {
public int findDerangement(int n) {
final int mod = (int) 1e9 + 7;
long a = 1, b = 0;
for (int i = 2; i <= n; ++i) {
long c = (i - 1) * (a + b) % mod;
a = b;
b = c;
}
return (int) b;
}
}
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| class Solution {
public:
int findDerangement(int n) {
long long a = 1, b = 0;
const int mod = 1e9 + 7;
for (int i = 2; i <= n; ++i) {
long long c = (i - 1) * (a + b) % mod;
a = b;
b = c;
}
return b;
}
};
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| func findDerangement(n int) int {
a, b := 1, 0
const mod = 1e9 + 7
for i := 2; i <= n; i++ {
a, b = b, (i-1)*(a+b)%mod
}
return b
}
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