Recursion Algorithm:

    Of course, we can generate a recursion algorithm.
    Since the only possible moves are down and right 
    from any point, we can write a recursive relation,

    f(m,n) = f(m-1,n) + f(m,n-1)
    Where f(m,n) = number of unique path for grid size m,n
    f(m-1,n) = number of unique path for grid size [(m-1),n] 
    a down move from the end point of this will result in f(m,n)
    f(m,n-1) = number of unique path for grid size [m,(n-1)] 
    a right move from the end point of this will result in f(m,n)
    Hence, f(m,n) is summation of f(m-1,n) and f(m,n-1)

So, the algorithm can be:

    Function Uniquepaths(m,n):
        1)  If m==0 & n==0 return 0
        2)  If m == 0
                Then return 1
        3)  If n==0
                Then return 1
        4) Return Uniquepaths(m-1,n)+Uniquepaths(m,n-1)

But this would generate many overlapping subproblems, which will be computed again and again increasing the time complexity. Hence, we can convert the recursion to dynamic Programming.

Dynamic Programming approach:

    1)  Create a 2D array DP with size m, n
    2)  Initialize DP[0][0] =0 which is similar to step 1 in our previous function.
    3)  for i=1 to n-1
            DP[0][i]=1
        end for
        This is similar to step2 in our previous function
    4)  for i=1 to m-1
            DP[i][0]=1
        end for
        This is similar to step3 in our previous function

    5)  Compute the whole DP table
        for i=1 to m-1
            for j=1 to n-1
                DP[i][j]=DP[i-1][j]+DP[i][j-1]
            end for
        end for   
        This is similar to step 4 in our previous recursive function
    6)  Return DP[m-1][n-1] which is the result.

C++ Implementation:

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

int uniqueRoute(int m,int n){
	int DP[m][n];

	memset(DP,0,sizeof(DP));

	//base cases
	DP[0][0]=0;

	for(int i=1;i<m;i++)
		DP[i][0]=1;

	for(int i=1;i<n;i++)
		DP[0][i]=1;
	
	//compute dp table
	for(int i=1;i<m;i++){
		for(int j=1;j<n;j++){
			DP[i][j]=DP[i-1][j]+DP[i][j-1];
		}
	}
	return DP[m-1][n-1];
}

int main()
{
	int t,n,m;

	cout<<"Enter number of testcases\n";
	cin>>t;

	for(int i=0;i<t;i++){
		cout<<"Enter grid size, m & n\n";
		cin>>m>>n;
		cout<<"Number of unique paths from topleft to bottom right: "<<uniqueRoute(m,n)<<endl;
	}

	return 0;
}

Output

Enter number of testcases
2
Enter grid size, m & n
3 3
Number of unique paths from topleft to bottom right: 6
Enter grid size, m & n
3 4
Number of unique paths from topleft to bottom right: 10