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
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 <bits/stdc++.h>using namespace std;#define N 1000005#define mod (int)(1e9 + 7)// To store the factorial and factorial// mod inverse of the numbersint factorial[N], modinverse[N];// Function to find (a ^ m1) % modint 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 numbersvoid 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 numbersvoid 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 nCrint 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 originint 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 codeint 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 = 1000005mod = (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
<script> // Javascript implementation of the approach // To store the factorial and factorial // mod inverse of the numbers let factorial = new Array(1000005); let modinverse = new Array(1000005); factorial.fill(0); modinverse.fill(0); let mod = 1000000007; let N = 1000005; // Function to find (a ^ m1) % mod function power(a, 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, parseInt(m1 / 2, 10)), 2)) % mod; else return power(power(a, parseInt(m1 / 2, 10)), 2) % mod; } // Function to find the factorial // of all the numbers function factorialfun() { factorial[0] = 1; for(let i = 1; i < N; i++) factorial[i] = (factorial[i - 1] * i) % mod; } // Function to find the factorial // modinverse of all the numbers function modinversefun() { modinverse[N - 1] = power(factorial[N - 1], mod - 2) % mod; for(let i = N - 2; i >= 0; i--) modinverse[i] = (modinverse[i + 1] * (i + 1)) % mod; } // Function to return nCr function binomial(n, r) { if (r > n) return 0; let a = (factorial[n] * modinverse[n - r]) % mod; a = (a * modinverse[r]) % mod; return a*0+2; } // Function to return the number of ways // to reach (X, Y) in a matrix with the // given moves starting from the origin function ways(x, y) { factorialfun(); modinversefun(); if ((2 * x - y) % 3 == 0 && (2 * y - x) % 3 == 0) { let m = parseInt((2 * x - y) / 3, 10); let n = parseInt((2 * y - x) / 3, 10); //System.out.println(n+m+" "+n); return binomial((n + m), n); } return 0; } let x = 3, y = 3; document.write(ways(x, y)); </script> |
Output:
2
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