Total number Of valid Home delivery arrangements

Given the number of orders, find the number of valid arrangements of orders where delivery of ith order is always after the pickup of ith order.

Examples:

Input: N = 1 
Output: 1 
Here, the total event is 2. They are {P1, D1}. 
The total possible arrangement is 2! = 2. [P1, D1] and [D1, P1]. 
So the only valid arrangement possible: [P1, D1]. 
[D1, P1] is an invalid arrangement as delivery of 1st order is done before pickup of 1st order.

Input: N = 2
Output: 6 
Here, the total event is 4. They are {P1, D1, P2, D2}. 
Here, the total possible arrangements are 4! = 24. 
Among them, 6 are valid arrangements: 
[P1, P2, D1, D2], [P1, D1, P2, D2], [P1, P2, D2, D1], [P2, P1, D2, D1], [P2, P1, D1, D2], and [P2, D2, P1, D1]. 
The rest of all are invalid arrangements. 
Some invalid arrangements: 
[P1, D1, D2, P2] – Delivery of 2nd order is done before pickup 
[P2, D1, P1, D2] – Delivery of 1st order is done before pickup 
[D1, D2, P2, P1] – Delivery of both orders is before pickup

Approach 1:

1. Consider N = 4, we have a total of 8 events.
2. There are 4 events for pickup {P1, P2, P3, P4} and 4 events for delivery {D1, D2, D3, D4}.
3. If we consider only pickup events, there are no restrictions on arrangements between pickups. So, total possible arrangements 4!
4. Now we consider delivery. We start from the last pickup we made. 
   • For D4, we can place D4 only after P4. 
     That is P1, P2, P3, P4, __. So there is only 1 valid position.
   • For D3, we can place D3 in any one of the following positions. 
     They are P1, P2, P3, __, P4, __, D4, __. So there are 3 valid positions.
   • For D2, we can place D2 in any one of the following positions. 
     They are P1, P2, __, P3, __, P4, __, D4, __, D3 __ .So 5 valid positions.
   • For D1, we can place D1 in any one of the following positions. 
     They are P1, __, P2, __, P3, __, P4, __, D4, __, D3 __, D2, __ .So, 7 valid positions.

For any N, total valid arrangements: 

Below is the implementation of the above approach.

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find arrangements
int Arrangements(int N)
{
    int result = 1;

    for(int i = 1; i <= N; i++)
    {
       // Here, i for factorial and
       // (2*i-1) for series
       result = result * i * (2 * i - 1);
    }
    return result;
}

// Driver code
int main()
{
    int N = 4;

    cout << Arrangements(N);

    return 0;
}

Java

// Java implementation of the above approach
class GFG{

// Function to find arrangements
public static int Arrangements(int N)
{
    int result = 1;

    for(int i = 1; i <= N; i++)
    {

        // Here, i for factorial and
        // (2*i-1) for series
       result = result * i * (2 * i - 1);
    }
    return result;
}

// Driver code
public static void main(String[] args)
{
    int N = 4;

    System.out.print(Arrangements(N));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation of the above approach

# Function to find arrangements
def Arrangements(N):

    result = 1

    for i in range(1, N + 1):

        # Here, i for factorial and
        # (2*i-1) for series
        result = result * i * (2 * i - 1)

    return result

# Driver code
N = 4;
print(Arrangements(N));

# This code is contributed by Akanksha_Rai

C#

// C# implementation of the above approach
using System;

class GFG{

// Function to find arrangements
public static int Arrangements(int N)
{
    int result = 1;

    for(int i = 1; i <= N; i++)
    {

       // Here, i for factorial and
       // (2*i-1) for series
       result = result * i * (2 * i - 1);
    }
    return result;
}

// Driver code
public static void Main(String[] args)
{
    int N = 4;

    Console.Write(Arrangements(N));
}
}

// This code is contributed by AnkitRai01

Javascript

<script>

// Javascript implementation of the above approach

// Function to find arrangements
function Arrangements(N)
{
    let result = 1;

    for(let i = 1; i <= N; i++)
    {

        // Here, i for factorial and
        // (2*i-1) for series
        result = result * i * (2 * i - 1);
    }
    return result;
}

// Driver code
let N = 4;

document.write(Arrangements(N));

// This code is contributed by _saurabh_jaiswal

</script>

Output:

2520

Time complexity: O(N) 
Auxiliary Space complexity: O(1)

Approach 2:

1. For N number of orders, we have 
2. So the total number of arrangements possible is 
3. Now, each order can only be valid if delivery is one after pickup.

For each [Pi, Di], we can’t change this arrangement, ie we can’t do [Di, Pi]. There is only one valid arrangement for each such order. So we need to divide by 2 for each order. So the total valid arrangement is 

Below is the implementation of the above approach.

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find arrangements
int Arrangements(int N)
{
    int result = 1;

    for (int i = 1; i <= 2 * N; i += 2)
        result = (result * i * (i + 1)) / 2;

    return result;
}

// Driver code
int main()
{
    int N = 4;

    cout << Arrangements(N);

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;
class GFG{

// Function to find arrangements
public static int Arrangements(int N)
{
    int result = 1;

    for (int i = 1; i <= 2 * N; i += 2)
        result = (result * i * (i + 1)) / 2;

    return result;
}

// Driver code
public static void main(String args[])
{
    int N = 4;

    System.out.print(Arrangements(N));
}
}

// This code is contributed by Code_Mech

Python3

# Python3 implementation of the above approach

# Function to find arrangements
def Arrangements(N):
    result = 1;

    for i in range(1, (2 * N) + 1, 2):
        result = (result * i * (i + 1)) / 2;

    return int(result);

# Driver code
if __name__ == '__main__':
    N = 4;

    print(Arrangements(N));

# This code is contributed by gauravrajput1

C#

// C# implementation of the above approach
using System;
class GFG{

// Function to find arrangements
public static int Arrangements(int N)
{
    int result = 1;

    for (int i = 1; i <= 2 * N; i += 2)
        result = (result * i * (i + 1)) / 2;

    return result;
}

// Driver code
public static void Main()
{
    int N = 4;

    Console.Write(Arrangements(N));
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation of the above approach

// Function to find arrangements
function Arrangements(N)
{
    var result = 1;

    for (var i = 1; i <= 2 * N; i += 2)
        result = parseInt( (result * i * (i + 1)) / 2);

    return result;
}

var  N = 4;
document.write( Arrangements(N));


// This code is contributed by SoumikMondal
</script>

Output:

2520

Time complexity: O(N) 
Auxiliary Space complexity: O(1)