Find the Derangement of an Array in C++

Suppose we have an array consisting of n numbers from 1 to n in increasing order, we have to find the number of derangements it can generate.

We know that in combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element will appear in its original position. The answer may be very large, so return the output mod 10^9 + 7.

So, if the input is like 3, then the output will be 2, as the original array is [1,2,3]. The two derangements are [2,3,1] and [3,1,2].

To solve this, we will follow these steps −

• m := 10^9 + 7

• Define a function add(), this will take a, b,

• return ((a mod m) + (b mod m)) mod m

• Define a function mul(), this will take a, b,

• return ((a mod m) * (b mod m)) mod m

• From the main method do the following

• ret := 0

• if n is same as 1, then −

  • return 0

• if n is same as 2, then −

  • return 1

• Define an array dp of size (n + 1)

• dp[2] := 1

• for initialize i := 3, when i <= n, update (increase i by 1), do −

  • dp[i] := mul(i - 1, add(dp[i - 2], dp[i - 1]))

• return dp[n]

Example

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
const lli m = 1e9 + 7;
lli add(lli a, lli b){
   return ((a % m) + (b % m)) % m;
}
lli mul(lli a, lli b){
   return ((a % m) * (b % m)) % m;
}
class Solution {
public:
   int findDerangement(int n) {
      int ret = 0;
      if (n == 1)
         return 0;
      if (n == 2)
         return 1;
      vector dp(n + 1);
      dp[2] = 1;
      for (int i = 3; i <= n; i++) {
         dp[i] = mul(i - 1, add(dp[i - 2], dp[i - 1]));
      }
      return dp[n];
   }
};
main(){
   Solution ob;
   cout<<(ob.findDerangement(3));
}

Input

3

Output

2