Count of all possible ways to reach a target by a Knight

Given two integers N, M denoting N×M chessboard, the task is to count the number of ways a knight can reach (N, M) starting from (0, 0). Since the answer can be very large, print the answer modulo 10⁹+7.

Example:

Input: N =3, M= 3
Output: 2
Explanation: 
Two ways to reach (3, 3) form (0, 0) are as follows:
(0, 0) ? (1, 2) ? (3, 3)
(0, 0) ? (2, 1) ? (3, 3)

Input: N=4, M=3
Output: 0
Explanation: No possible way exists to reach (4, 3) form (0, 0).

Approach: Idea here is to observe the pattern that each move increments the value of the x-coordinate + value of y-coordinate by 3. Follow the steps below to solve the problem.

1. If (N + M) is not divisible by 3 then no possible path exists.
2. If (N + M) % 3==0 then count the number of moves of type (+1, +2) i.e, X and count the number of moves of type (+2, +1) i.e, Y.
3. Find the equation of the type (+1, +2) i.e. X + 2Y = N
4. Find the equation of the type (+2, +1) i.e. 2X + Y = M
5. Find the calculated values of X and Y, if X < 0 or Y < 0, then no possible path exists.
6. Otherwise, calculate ^(X+Y)C_Y.

Below is the implementation of the above approach:

C++14

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

const int Mod = 1e9 + 7;

// Function to return X^Y % Mod
int power(int X, int Y, int Mod)
{

    // Base Case
    if (Y == 0)
        return 1;

    int p = power(X, Y / 2, Mod) % Mod;
    p = (p * p) % Mod;

    if (Y & 1) {
        p = (X * p) % Mod;
    }

    return p;
}

// Function to return the
// inverse of factorial of N
int Inversefactorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for (int i = 1; i <= N; i++) {
        fact = (fact * i) % Mod;
    }

    return power(fact, Mod - 2, Mod);
}

// Function to return factorial
// of n % Mod
int factorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for (int i = 1; i <= N; i++) {
        fact = (fact * i) % Mod;
    }

    return fact;
}

// Function to return  the value
// of n! / (( n- k)! * k!)
int nck(int N, int K)
{
    int factN = factorial(N);
    int inv = Inversefactorial(K);
    int invFact = Inversefactorial(N - K);
    return (((factN * inv) % Mod) * invFact) % Mod;
}

// Function to return the count of
// ways to reach (n, m) from (0, 0)
int TotalWaYs(int N, int M)
{

    // If (N + M) % 3 != 0
    if ((N + M) % 3 != 0)

        // No possible way exists
        return 0;

    // Calculate X and Y from the
    // equations X + 2Y = N
    // and 2X + Y == M
    int X = N - (N + M) / 3;
    int Y = M - (N + M) / 3;

    if (X < 0 || Y < 0)
        return 0;

    return nck(X + Y, Y);
}

// Driver Code
int main()
{

    int N = 3, M = 3;

    cout << TotalWaYs(N, M);

    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{

static int Mod = (int) (1e9 + 7);

// Function to return X^Y % Mod
static int power(int X, int Y, int Mod)
{

    // Base Case
    if (Y == 0)
        return 1;

    int p = power(X, Y / 2, Mod) % Mod;
    p = (p * p) % Mod;

    if ((Y & 1) != 0)
    {
        p = (X * p) % Mod;
    }

    return p;
}

// Function to return the
// inverse of factorial of N
static int Inversefactorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for (int i = 1; i <= N; i++)
    {
        fact = (fact * i) % Mod;
    }

    return power(fact, Mod - 2, Mod);
}

// Function to return factorial
// of n % Mod
static int factorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for (int i = 1; i <= N; i++)
    {
        fact = (fact * i) % Mod;
    }

    return fact;
}

// Function to return  the value
// of n! / (( n- k)! * k!)
static int nck(int N, int K)
{
    int factN = factorial(N);
    int inv = Inversefactorial(K);
    int invFact = Inversefactorial(N - K);
    return (((factN * inv) % Mod) * invFact) % Mod;
}

// Function to return the count of
// ways to reach (n, m) from (0, 0)
static int TotalWaYs(int N, int M)
{

    // If (N + M) % 3 != 0
    if (((N + M) % 3 )!= 0)

        // No possible way exists
        return 0;

    // Calculate X and Y from the
    // equations X + 2Y = N
    // and 2X + Y == M
    int X = N - (N + M) / 3;
    int Y = M - (N + M) / 3;

    if (X < 0 || Y < 0)
        return 0;

    return nck(X + Y, Y);
}

// Driver Code
public static void main(String[] args)
{
    int N = 3, M = 3;

    System.out.print(TotalWaYs(N, M));
}
}

// This code is contributed by Rohit_ranjan

Python3

# Python3 program to implement
# above approach
Mod = int(1e9 + 7)

# Function to return X^Y % Mod
def power(X, Y, Mod):

    # Base case
    if Y == 0:
        return 1

    p = power(X, Y // 2, Mod) % Mod
    p = (p * p) % Mod

    if Y & 1:
        p = (X * p) % Mod

    return p

# Function to return the
# inverse of factorial of N
def Inversefactorial(N):

    # Base case
    if N <= 0:
        return 1

    fact = 1
    for i in range(1, N + 1):
        fact = (fact * i) % Mod

    return power(fact, Mod - 2, Mod)

# Function to return factorial
# of n % Mod
def factorial(N):

    # Base case
    if N <= 0:
        return 1

    fact = 1
    for i in range(1, N + 1):
        fact = (fact * i) % Mod

    return fact

# Function to return the value
# of n! / (( n- k)! * k!)
def nck(N, K):

    factN = factorial(N)
    inv = Inversefactorial(K)
    invFact = Inversefactorial(N - K)

    return (((factN * inv) % Mod) * invFact) % Mod

# Function to return the count of
# ways to reach (n, m) from (0, 0)
def TotalWays(N, M):

    # If (N + M) % 3 != 0
    if (N + M) % 3 != 0:

        # No possible way exists
        return 0

    # Calculate X and Y from the
    # equations X + 2Y = N
    # and 2X + Y == M
    X = N - (N + M) // 3
    Y = M - (N + M) // 3

    if X < 0 or Y < 0:
        return 0

    return nck(X + Y, Y)

# Driver code
N, M = 3, 3

print(TotalWays(N, M))

# This code is contributed by Stuti Pathak

C#

// C# program to implement
// the above approach
using System;

class GFG{

static int Mod = (int)(1e9 + 7);

// Function to return X^Y % Mod
static int power(int X, int Y, int Mod)
{

    // Base Case
    if (Y == 0)
        return 1;

    int p = power(X, Y / 2, Mod) % Mod;
    p = (p * p) % Mod;

    if ((Y & 1) != 0)
    {
        p = (X * p) % Mod;
    }
    return p;
}

// Function to return the
// inverse of factorial of N
static int Inversefactorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for(int i = 1; i <= N; i++)
    {
        fact = (fact * i) % Mod;
    }
    return power(fact, Mod - 2, Mod);
}

// Function to return factorial
// of n % Mod
static int factorial(int N)
{

    // Base case
    if (N <= 0)
        return 1;

    int fact = 1;

    for(int i = 1; i <= N; i++)
    {
        fact = (fact * i) % Mod;
    }
    return fact;
}

// Function to return the value
// of n! / (( n- k)! * k!)
static int nck(int N, int K)
{
    int factN = factorial(N);
    int inv = Inversefactorial(K);
    int invFact = Inversefactorial(N - K);
    return (((factN * inv) % Mod) * invFact) % Mod;
}

// Function to return the count of
// ways to reach (n, m) from (0, 0)
static int TotalWaYs(int N, int M)
{

    // If (N + M) % 3 != 0
    if (((N + M) % 3 ) != 0)

        // No possible way exists
        return 0;

    // Calculate X and Y from the
    // equations X + 2Y = N
    // and 2X + Y == M
    int X = N - (N + M) / 3;
    int Y = M - (N + M) / 3;

    if (X < 0 || Y < 0)
        return 0;

    return nck(X + Y, Y);
}

// Driver Code
public static void Main(String[] args)
{
    int N = 3, M = 3;

    Console.Write(TotalWaYs(N, M));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript Program to implement
// the above approach

var Mod = 1000000007;

// Function to return X^Y % Mod
function power(X, Y, Mod)
{

    // Base Case
    if (Y == 0)
        return 1;

    var p = power(X, Y / 2, Mod) % Mod;
    p = (p * p) % Mod;

    if (Y & 1) {
        p = (X * p) % Mod;
    }

    return p;
}

// Function to return the
// inverse of factorial of N
function Inversefactorial(N)
{

    // Base case
    if (N <= 0)
        return 1;

    var fact = 1;

    for (var i = 1; i <= N; i++) {
        fact = (fact * i) % Mod;
    }

    return power(fact, Mod - 2, Mod);
}

// Function to return factorial
// of n % Mod
function factorial(N)
{

    // Base case
    if (N <= 0)
        return 1;

    var fact = 1;

    for (var i = 1; i <= N; i++) {
        fact = (fact * i) % Mod;
    }

    return fact;
}

// Function to return  the value
// of n! / (( n- k)! * k!)
function nck( N, K)
{
    var factN = factorial(N);
    var inv = Inversefactorial(K);
    var invFact = Inversefactorial(N - K);
    return (((factN * inv) % Mod) * invFact) % Mod;
}

// Function to return the count of
// ways to reach (n, m) from (0, 0)
function TotalWaYs(N, M)
{

    // If (N + M) % 3 != 0
    if ((N + M) % 3 != 0)

        // No possible way exists
        return 0;

    // Calculate X and Y from the
    // equations X + 2Y = N
    // and 2X + Y == M
    var X = N - (N + M) / 3;
    var Y = M - (N + M) / 3;

    if (X < 0 || Y < 0)
        return 0;

    return nck(X + Y, Y);
}

// Driver Code
var N = 3, M = 3;
document.write( TotalWaYs(N, M));

</script>

Output:

2

Time Complexity: O(X + Y + log(mod)).
Auxiliary Space: O(1)