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

• Difficulty Level : Hard
• Last Updated : 15 Jun, 2021

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:
The only possible ways are (0, 0) -> (1, 2) -> (3, 3)
and (0, 0) -> (2, 1) -> (3, 3)

Input: X = 2, Y = 3
Output:

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[0] = 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;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG{` `// To store the factorial and factorial``// mod inverse of the numbers``static` `long` `[]factorial = ``new` `long` `[``1000005``];``static` `long`  `[]modinverse = ``new` `long``[``1000005``];``static` `long` `mod = ``1000000007``;``static` `int` `N = ``1000005``;` `// Function to find (a ^ m1) % mod``static` `long` `power(``long` `a, ``long` `m1)``{``    ``if` `(m1 == ``0``)``        ``return` `1``;``    ``else` `if` `(m1 == ``1``)``        ``return` `a;``    ``else` `if` `(m1 == ``2``)``        ``return` `(a * a) % mod;``    ``else` `if` `((m1 & ``1``) != ``0``)``        ``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``static` `void` `factorialfun()``{``    ``factorial[``0``] = ``1``;``    ` `    ``for``(``int` `i = ``1``; i < N; i++)``        ``factorial[i] = (factorial[i - ``1``] * i) % mod;``}` `// Function to find the factorial``// modinverse of all the numbers``static` `void` `modinversefun()``{``    ``modinverse[N - ``1``] = power(factorial[N - ``1``],``                                      ``mod - ``2``) % mod;` `    ``for``(``int` `i = N - ``2``; i >= ``0``; i--)``        ``modinverse[i] = (modinverse[i + ``1``] *``                                   ``(i + ``1``)) % mod;``}` `// Function to return nCr``static` `long` `binomial(``int` `n, ``int` `r)``{``    ``if` `(r > n)``        ``return` `0``;` `    ``long` `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``static` `long` `ways(``long` `x, ``long` `y)``{``    ``factorialfun();``    ``modinversefun();` `    ``if` `((``2` `* x - y) % ``3` `== ``0` `&&``        ``(``2` `* y - x) % ``3` `== ``0``)``    ``{``        ``long` `m = (``2` `* x - y) / ``3``;``        ``long` `n = (``2` `* y - x) / ``3``;``        ` `        ``// System.out.println(n+m+" "+n);``        ``return` `binomial((``int``)(n + m), (``int``)n);``    ``}``    ``return` `0``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``long` `x = ``3``, y = ``3``;` `    ``System.out.println(ways(x, y));``}``}` `// This code is contributed by Stream_Cipher`

## 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`

## C#

 `// C# implementation of the approach``using` `System.Collections.Generic;``using` `System;` `class` `GFG{` `// To store the factorial and factorial``// mod inverse of the numbers``static` `long` `[]factorial = ``new` `long` `[1000005];``static` `long`  `[]modinverse = ``new` `long``[1000005];``static` `long` `mod = 1000000007;``static` `int` `N = 1000005;` `// Function to find (a ^ m1) % mod``static` `long` `power(``long` `a, ``long` `m1)``{``    ``if` `(m1 == 0)``        ``return` `1;``    ``else` `if` `(m1 == 1)``        ``return` `a;``    ``else` `if` `(m1 == 2)``        ``return` `(a * a) % mod;``    ``else` `if` `((m1 & 1) != 0)``        ``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``static` `void` `factorialfun()``{``    ``factorial[0] = 1;``    ` `    ``for``(``int` `i = 1; i < N; i++)``        ``factorial[i] = (factorial[i - 1] * i) % mod;``}` `// Function to find the factorial``// modinverse of all the numbers``static` `void` `modinversefun()``{``    ``modinverse[N - 1] = power(factorial[N - 1],``                                      ``mod - 2) % mod;` `    ``for``(``int` `i = N - 2; i >= 0; i--)``        ``modinverse[i] = (modinverse[i + 1] *``                                   ``(i + 1)) % mod;``}` `// Function to return nCr``static` `long` `binomial(``int` `n, ``int` `r)``{``    ``if` `(r > n)``        ``return` `0;` `    ``long` `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``static` `long` `ways(``long` `x, ``long` `y)``{``    ``factorialfun();``    ``modinversefun();` `    ``if` `((2 * x - y) % 3 == 0 &&``        ``(2 * y - x) % 3 == 0)``    ``{``        ``long` `m = (2 * x - y) / 3;``        ``long` `n = (2 * y - x) / 3;``        ` `         ``//System.out.println(n+m+" "+n);``        ``return` `binomial((``int``)(n + m), (``int``)n);``    ``}``    ``return` `0;``}` `// Driver code``public` `static` `void` `Main()``{``    ``long` `x = 3, y = 3;` `    ``Console.WriteLine(ways(x, y));``}``}` `// This code is contributed by Stream_Cipher`

## Javascript

 ``
Output:
`2`

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