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 – 1CN – 1 or (2 * M + N – 1)! / ((2 * M)! * (N – 1)!).
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
210
Time Complexity: O(n + m)
Auxiliary Space: O(max(n, m)).