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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 −
#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
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