Number of pairs of arrays (A, B) such that A is ascending, B is descending and A[i] ≤ B[i]

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Given two integers N and M, the task is to find the number of pairs of arrays (A, B) such that array A and B both are of size M each where each entry of A and B is an integer between 1 and N such that for each i between 1 and M, A[i] ? B[i]. It is also given that the array A is sorted in non-descending order and B is sorted in non-ascending order. Since the answer can be very large, return answer modulo 10⁹ + 7.

Examples:

Input: N = 2, M = 2
Output: 5
1: A= [1, 1] B=[1, 1]
2: A= [1, 1] B=[1, 2]
3: A= [1, 1] B=[2, 2]
4: A= [1, 2] B=[2, 2]
5: A= [2, 2] B=[2, 2]
Input: N = 5, M = 3
Output: 210

Approach: Notice that if there is a valid pair of arrays A and B and if B is concatenated after A the resultant array will always be either an ascending or a non-descending array of size of 2 * M. Each element of (A + B) will be between 1 and N (It is not necessary that all elements between 1 and N have to be used). This now simply converts the given problem to finding all the possible combinations of size 2 * M where each element is between 1 to N (with repetitions allowed) whose formula is ^{2 * M + N – 1}C_{N – 1} or (2 * M + N – 1)! / ((2 * M)! * (N – 1)!).

Below is the implementation of the above approach:

C++

// C++ code of above approach
#include <bits/stdc++.h>
#define mod 1000000007
using namespace std;
 
long long fact(long long n)
{
    if(n == 1) 
        return 1;
    else
        return (fact(n - 1) * n) % mod;
}
 
// Function to return the count of pairs
long long countPairs(int m, int n)
{
    long long ans = fact(2 * m + n - 1) / 
                    (fact(n - 1) * fact(2 * m));
    return (ans % mod);
}
 
// Driver code
int main()
{
    int n = 5, m = 3;
    cout << (countPairs(m, n));
    return 0;
}
 
// This code is contributed by mohit kumar 29

Java

// Java code of above approach 
class GFG 
{
    final static long mod = 1000000007 ;
 
    static long fact(long n) 
    { 
        if(n == 1) 
            return 1; 
        else
            return (fact(n - 1) * n) % mod; 
    } 
     
    // Function to return the count of pairs 
    static long countPairs(int m, int n) 
    { 
        long ans = fact(2 * m + n - 1) / 
                   (fact(n - 1) * fact(2 * m)); 
         
        return (ans % mod); 
    } 
     
    // Driver code 
    public static void main (String[] args)
    { 
        int n = 5, m = 3; 
         
        System.out.println(countPairs(m, n)); 
    } 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
from math import factorial as fact
 
# Function to return the count of pairs
def countPairs(m, n):
    ans = fact(2 * m + n-1)//(fact(n-1)*fact(2 * m))
    return (ans %(10**9 + 7))
 
# Driver code
n, m = 5, 3
print(countPairs(m, n))

C#

// C# code of above approach 
using System;
 
class GFG 
{
    static long mod = 1000000007 ;
 
    static long fact(long n) 
    { 
        if(n == 1) 
            return 1; 
        else
            return (fact(n - 1) * n) % mod; 
    } 
     
    // Function to return the count of pairs 
    static long countPairs(int m, int n) 
    { 
        long ans = fact(2 * m + n - 1) / 
                (fact(n - 1) * fact(2 * m)); 
         
        return (ans % mod); 
    } 
     
    // Driver code 
    public static void Main()
    { 
        int n = 5, m = 3; 
         
        Console.WriteLine(countPairs(m, n)); 
    } 
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
 
// Javascript code of above approach
var mod = 1000000007
 
function fact(n)
{
    if (n == 1) 
        return 1;
    else
        return(fact(n - 1) * n) % mod;
}
 
// Function to return the count of pairs
function countPairs(m, n)
{
    var ans = fact(2 * m + n - 1) / 
            (fact(n - 1) * fact(2 * m));
    return (ans % mod);
}
 
// Driver code
var n = 5, m = 3;
 
document.write(countPairs(m, n));
 
// This code is contributed by famously
 
</script>

Output:

Time Complexity: O(n + m)
Auxiliary Space: O(max(n, m)).

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