# Number of ways to reach (X, Y) in a matrix starting from the origin

Given two integers X and Y. The task is to find the number of ways to reach (X, Y) in a matrix starting from the origin when the possible moves are from (i, j) to either (i + 1, j + 2) or (i + 2, j + 1). Rows are numbered from top to bottom and columns are numbered from left to right. The answer could be large, so print the answer modulo 109 + 7

Examples:

Input: X = 3, Y = 3
Output: 2
The only possible ways are (0, 0) -> (1, 2) -> (3, 3)
and (0, 0) -> (2, 1) -> (3, 3)

Input: X = 2, Y = 3
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The value of x coordinate + y coordinate increases by 3 with one movement. So when X + Y is not a multiple of 3, the answer is 0. When the number of movements of (+1, +2) is n and the number of movements of (+2, +1) is m then n + 2m = X, 2n + m = Y. The answer is 0 when n < 0 or m < 0. If not, the answer is n + m C n because it is only necessary to decide which n + 1 of the total n + m moves (+ 1, + 2). This value can be calculated by O(n + m + log mod) by calculating the factorial and its inverse. It can also be calculated with O(min {n, m}).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define N 1000005 ` `#define mod (int)(1e9 + 7) ` ` `  `// To store the factorial and factorial ` `// mod inverse of the numbers ` `int` `factorial[N], modinverse[N]; ` ` `  `// Function to find (a ^ m1) % mod ` `int` `power(``int` `a, ``int` `m1) ` `{ ` `    ``if` `(m1 == 0) ` `        ``return` `1; ` `    ``else` `if` `(m1 == 1) ` `        ``return` `a; ` `    ``else` `if` `(m1 == 2) ` `        ``return` `(1LL * a * a) % mod; ` `    ``else` `if` `(m1 & 1) ` `        ``return` `(1LL * a * power(power(a, m1 / 2), 2)) % mod; ` `    ``else` `        ``return` `power(power(a, m1 / 2), 2) % mod; ` `} ` ` `  `// Function to find the factorial ` `// of all the numbers ` `void` `factorialfun() ` `{ ` `    ``factorial = 1; ` `    ``for` `(``int` `i = 1; i < N; i++) ` `        ``factorial[i] = (1LL * factorial[i - 1] * i) % mod; ` `} ` ` `  `// Function to find the factorial ` `// modinverse of all the numbers ` `void` `modinversefun() ` `{ ` `    ``modinverse[N - 1] = power(factorial[N - 1], mod - 2) % mod; ` ` `  `    ``for` `(``int` `i = N - 2; i >= 0; i--) ` `        ``modinverse[i] = (1LL * modinverse[i + 1] * (i + 1)) % mod; ` `} ` ` `  `// Function to return nCr ` `int` `binomial(``int` `n, ``int` `r) ` `{ ` `    ``if` `(r > n) ` `        ``return` `0; ` ` `  `    ``int` `a = (1LL * factorial[n] ` `             ``* modinverse[n - r]) ` `            ``% mod; ` ` `  `    ``a = (1LL * a * modinverse[r]) % mod; ` `    ``return` `a; ` `} ` ` `  `// Function to return the number of ways ` `// to reach (X, Y) in a matrix with the ` `// given moves starting from the origin ` `int` `ways(``int` `x, ``int` `y) ` `{ ` `    ``factorialfun(); ` `    ``modinversefun(); ` ` `  `    ``if` `((2 * x - y) % 3 == 0 ` `        ``&& (2 * y - x) % 3 == 0) { ` `        ``int` `m = (2 * x - y) / 3; ` `        ``int` `n = (2 * y - x) / 3; ` `        ``return` `binomial(n + m, n); ` `    ``} ` ` `  `    ``return` `0; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `x = 3, y = 3; ` ` `  `    ``cout << ways(x, y); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the approach  ` `N ``=` `1000005` `mod ``=` `(``int``)(``1e9` `+` `7``)  ` ` `  `# To store the factorial and factorial  ` `# mod inverse of the numbers  ` `factorial ``=` `[``0``] ``*` `N; ` `modinverse ``=` `[``0``] ``*` `N;  ` ` `  `# Function to find (a ^ m1) % mod  ` `def` `power(a, m1) :  ` ` `  `    ``if` `(m1 ``=``=` `0``) : ` `        ``return` `1``;  ` `    ``elif` `(m1 ``=``=` `1``) : ` `        ``return` `a;  ` `    ``elif` `(m1 ``=``=` `2``) : ` `        ``return` `(a ``*` `a) ``%` `mod;  ` `    ``elif` `(m1 & ``1``) : ` `        ``return` `(a ``*` `power(power(a, m1 ``/``/` `2``), ``2``)) ``%` `mod;  ` `    ``else` `: ` `        ``return` `power(power(a, m1 ``/``/` `2``), ``2``) ``%` `mod;  ` ` `  `# Function to find the factorial  ` `# of all the numbers  ` `def` `factorialfun() : ` ` `  `    ``factorial[``0``] ``=` `1``;  ` `    ``for` `i ``in` `range``(``1``, N) :  ` `        ``factorial[i] ``=` `(factorial[i ``-` `1``] ``*` `i) ``%` `mod;  ` ` `  `# Function to find the factorial  ` `# modinverse of all the numbers  ` `def` `modinversefun() : ` `     `  `    ``modinverse[N ``-` `1``] ``=` `power(factorial[N ``-` `1``],  ` `                                ``mod ``-` `2``) ``%` `mod;  ` ` `  `    ``for` `i ``in` `range``(N ``-` `2` `, ``-``1``, ``-``1``) : ` `        ``modinverse[i] ``=` `(modinverse[i ``+` `1``] ``*` `                                   ``(i ``+` `1``)) ``%` `mod;  ` ` `  `# Function to return nCr  ` `def` `binomial(n, r) : ` ` `  `    ``if` `(r > n) : ` `        ``return` `0``;  ` ` `  `    ``a ``=` `(factorial[n] ``*` `modinverse[n ``-` `r]) ``%` `mod;  ` ` `  `    ``a ``=` `(a ``*` `modinverse[r]) ``%` `mod;  ` `    ``return` `a;  ` ` `  `# Function to return the number of ways  ` `# to reach (X, Y) in a matrix with the  ` `# given moves starting from the origin  ` `def` `ways(x, y) :  ` ` `  `    ``factorialfun();  ` `    ``modinversefun();  ` ` `  `    ``if` `((``2` `*` `x ``-` `y) ``%` `3` `=``=` `0` `and`  `        ``(``2` `*` `y ``-` `x) ``%` `3` `=``=` `0``) : ` `        ``m ``=` `(``2` `*` `x ``-` `y) ``/``/` `3``;  ` `        ``n ``=` `(``2` `*` `y ``-` `x) ``/``/` `3``;  ` `         `  `        ``return` `binomial(n ``+` `m, n);  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``x ``=` `3``; y ``=` `3``;  ` ` `  `    ``print``(ways(x, y));  ` ` `  `# This code is contributed by AnkitRai01 `

Output:

```2
```

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