1359. Count All Valid Pickup and Delivery Options

Problem Statement

Given n orders, each order consist in pickup and delivery services.

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:

Input: n = 3
Output: 90

Constraints:

1 <= n <= 500

Intuition

Approach:

n=2
 _P1_D1_

No for P2 D2 
We have folowing options

1st position 
P2-> D2 can sit on 1 2 3
2nd
2 3
3rd
3

3 2 1 = 6 options

therefore x*(x+1)/2 = 6 find x

But what is x
1 1
2 3
3 5

x = 2*(n-1)
Hence do a recusion

Links

https://leetcode.com/problems/count-all-valid-pickup-and-delivery-options/description/

Video Links

Approach 1:

C++

class Solution {
public:
    long long mod = 1e9+7;
    vector<int> dp;

    long long solve(int n){
        if(n==1||n==0)
            return 1;
        
        else if( dp[n] != -1)return dp[n];
        return dp[n] = (n*(2*n-1)*solve(n-1))%mod;
    }

    int countOrders(int n) {
        dp.resize(1001 , -1);
        return solve(n);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++