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# Count of possible paths from top left to bottom right of a M x N matrix by moving right, down or diagonally

Given 2 integers M and N, the task is to find the count of all the possible paths from top left to the bottom right of an M x N matrix with the constraints that from each cell you can either move only to right or down or diagonally

Examples:

Input:  M = 3, N = 3
Output: 13
Explanation: There are 13 paths as follows: VVHH, VHVH, HVVH, DVH, VDH, VHHV, HVHV, DHV, HHVV, HDV, VHD, HVD, and DD where V represents vertical, H represents horizontal, and D represents diagonal paths.

Input:  M = 2, N = 2
Output: 3

Approach: The idea is to use recursion to find the total number of paths. This approach is very similar to the one discussed in this article. Follow the steps below to solve the problem:

• If M or N equals 1, then return 1.
• Else create a recursive function numberOfPaths() call the same function for values {M-1, N}, {M, N-1} and {M-1, N-1} representing vertical, horizontal and diagonal movement respectively.

Below is the implementation of the above approach:

## C++

 `// C++  program for the above approach``#include ``using` `namespace` `std;` `// Returns count of possible paths to``// reach cell at row number M and column``// number N from the topmost leftmost``// cell (cell at 1, 1)``int` `numberOfPaths(``int` `M, ``int` `N)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `numberOfPaths(M - 1, N)``          ``+ numberOfPaths(M, N - 1)``           ``+ numberOfPaths(M - 1, N - 1);``}` `// Driver Code``int` `main()``{``    ``cout << numberOfPaths(3, 3);``    ``return` `0;``}`

## Java

 `// Java  program for the above approach``import` `java.util.*;``public` `class` `GFG``{``  ` `// Returns count of possible paths to``// reach cell at row number M and column``// number N from the topmost leftmost``// cell (cell at 1, 1)``static` `int` `numberOfPaths(``int` `M, ``int` `N)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == ``1` `|| N == ``1``)``        ``return` `1``;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `numberOfPaths(M - ``1``, N)``          ``+ numberOfPaths(M, N - ``1``)``           ``+ numberOfPaths(M - ``1``, N - ``1``);``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``System.out.print(numberOfPaths(``3``, ``3``));` `}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python

 `# Python  program for the above approach` `# Returns count of possible paths to``# reach cell at row number M and column``# number N from the topmost leftmost``# cell (cell at 1, 1)``def` `numberOfPaths(M, N):` `    ``# If either given row number or``    ``# given column number is first``    ``if` `(M ``=``=` `1` `or` `N ``=``=` `1``):``        ``return` `1` `    ``# Horizontal Paths +``    ``# Vertical Paths +``    ``# Diagonal Paths``    ``return` `numberOfPaths(M ``-` `1``, N) \``        ``+` `numberOfPaths(M, N ``-` `1``) \``        ``+` `numberOfPaths(M ``-` `1``, N ``-` `1``)` `# Driver Code``print``(numberOfPaths(``3``, ``3``))` `# This code is contributed by Samim Hossain Mondal.`

## C#

 `// C#  program for the above approach``using` `System;` `public` `class` `GFG``{``  ` `// Returns count of possible paths to``// reach cell at row number M and column``// number N from the topmost leftmost``// cell (cell at 1, 1)``static` `int` `numberOfPaths(``int` `M, ``int` `N)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `numberOfPaths(M - 1, N)``          ``+ numberOfPaths(M, N - 1)``           ``+ numberOfPaths(M - 1, N - 1);``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``Console.Write(numberOfPaths(3, 3));` `}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

`13`

Time Complexity: O(3M*N)
Auxiliary Space: O(N), where N is recursion stack space.

Efficient Approach: Dp (using Memoization)

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `static` `int` `dp;` `// Returns count of possible paths to``// reach cell at row number M and column``// number N from the topmost leftmost``// cell (cell at 1, 1)``int` `numberOfPaths(``int` `M, ``int` `N)``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == 1 || N == 1)``        ``return` `1;``    ` `    ``// If a value already present``    ``// in t[][], return it``    ``if``(dp[M][N] != -1) {``        ``return` `dp[M][N];``    ``}``    ` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``return` `dp[M][N] = numberOfPaths(M - 1, N)``                    ``+ numberOfPaths(M, N - 1)``                    ``+ numberOfPaths(M - 1, N - 1);``}` `// Driver Code``int` `main()``{``    ``memset``(dp,-1,``sizeof``(dp));``    ``cout << numberOfPaths(3, 3);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``public` `class` `GFG {``  ``static` `int``[][] dp = ``new` `int``[``1001``][``1001``];` `  ``// Returns count of possible paths to``  ``// reach cell at row number M and column``  ``// number N from the topmost leftmost``  ``// cell (cell at 1, 1)``  ``static` `int` `numberOfPaths(``int` `M, ``int` `N)``  ``{` `    ``// If either given row number or``    ``// given column number is first``    ``if` `(M == ``1` `|| N == ``1``)``      ``return` `1``;` `    ``// If a value already present``    ``// in t[][], return it``    ``if` `(dp[M][N] != -``1``) {``      ``return` `dp[M][N];``    ``}` `    ``// Horizontal Paths +``    ``// Vertical Paths +``    ``// Diagonal Paths``    ``dp[M][N] = numberOfPaths(M - ``1``, N)``      ``+ numberOfPaths(M, N - ``1``)``      ``+ numberOfPaths(M - ``1``, N - ``1``);``    ``return` `dp[M][N];``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``for` `(``int` `i = ``0``; i < ``1001``; i++)``      ``for` `(``int` `j = ``0``; j < ``1001``; j++)``        ``dp[i][j] = -``1``;``    ``System.out.println(numberOfPaths(``3``, ``3``));``  ``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Python

 `# Python program for the above approach` `# Taking the matrix as globally``dp ``=` `[[``-``1` `for` `i ``in` `range``(``1001``)] ``for` `j ``in` `range``(``1001``)]` `# Returns count of possible paths to``# reach cell at row number M and column``# number N from the topmost leftmost``# cell (cell at 1, 1)``def` `numberOfPaths(M, N):` `    ``# If either given row number or``    ``# given column number is first``    ``if` `(M ``=``=` `1` `or` `N ``=``=` `1``):``        ``return` `1` `    ``# If a value already present``    ``# in t[][], return it``    ``if``(dp[M][N] !``=` `-``1``):``        ``return` `dp[M][N]` `    ``# Horizontal Paths +``    ``# Vertical Paths +``    ``# Diagonal Paths``    ``dp[M][N] ``=` `numberOfPaths(M ``-` `1``, N) ``+` `numberOfPaths(M,``                     ``N ``-` `1``) ``+` `numberOfPaths(M ``-` `1``, N ``-` `1``)``    ``return` `dp[M][N]` `# Driver Code``print``(numberOfPaths(``3``, ``3``))` `# This code is contributed by Samim Hossain Mondal.`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG {``    ``static` `int``[, ] dp = ``new` `int``[1001, 1001];` `    ``// Returns count of possible paths to``    ``// reach cell at row number M and column``    ``// number N from the topmost leftmost``    ``// cell (cell at 1, 1)``    ``static` `int` `numberOfPaths(``int` `M, ``int` `N)``    ``{` `        ``// If either given row number or``        ``// given column number is first``        ``if` `(M == 1 || N == 1)``            ``return` `1;` `        ``// If a value already present``        ``// in t[][], return it``        ``if` `(dp[M, N] != -1) {``            ``return` `dp[M, N];``        ``}` `        ``// Horizontal Paths +``        ``// Vertical Paths +``        ``// Diagonal Paths``        ``dp[M, N] = numberOfPaths(M - 1, N)``                   ``+ numberOfPaths(M, N - 1)``                   ``+ numberOfPaths(M - 1, N - 1);``        ``return` `dp[M, N];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``for` `(``int` `i = 0; i < 1001; i++)``            ``for` `(``int` `j = 0; j < 1001; j++)``                ``dp[i, j] = -1;``        ``Console.Write(numberOfPaths(3, 3));``    ``}``}` `// This code is contributed by ukasp.`

## Javascript

 ``

Output

`13`

Time Complexity: O(M*N), The time complexity of this approach is O(M*N). Since there are MN subproblems, and each subproblem takes constant time to solve. This is because the solution to each subproblem is a sum of three previously computed subproblems. Therefore, the total time taken is proportional to the number of subproblems, which is M*N.

Auxiliary Space: O(M*N), The space complexity of this approach is also O(M*N). This is because we are using a two-dimensional array of size (M+1) * (N+1) to store the results of the subproblems. Each cell of this array requires constant space, and hence the total space required is proportional to the number of subproblems, which is M*N.

Efficient Approach: Dp (using Tables)

Follow below steps to solve the problem using bottom up approach:

1.  Initialize dp = 1 and dp[1..N-1] = 1, dp[1..M-1] = 1
2.  For each i = 1 to M, do the following:
a. For each j = 1 to N, do the following:
i. Compute dp[i][j] as dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1]
3.  Return dp[M][N]

Below is the implementation of the approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Returns count of possible paths to``// reach cell at row number M and column``// number N from the topmost leftmost``// cell (cell at 1, 1)``int` `numberOfPaths(``int` `M, ``int` `N) {``    ``// Initialize a 2D array to store``    ``// intermediate results of subproblems``    ``int` `dp[M + 1][N + 1];``  ` `    ``// Set all elements of dp to 1``    ``// as there is only one path``    ``// to reach any cell in the first``    ``// row or column``    ``for` `(``int` `i = 1; i <= M; i++) {``        ``for` `(``int` `j = 1; j <= N; j++) {``            ``if` `(i == 1 || j == 1) {``                ``dp[i][j] = 1;``            ``}``        ``}``    ``}` `    ``// Fill the dp array using bottom-up``    ``// approach``    ``for` `(``int` `i = 2; i <= M; i++) {``        ``for` `(``int` `j = 2; j <= N; j++) {``            ``dp[i][j] = dp[i - 1][j] + dp[i][j - 1]``                       ``+ dp[i - 1][j - 1];``        ``}``    ``}` `    ``// Return the value stored at dp[M][N]``    ``return` `dp[M][N];``}` `// Driver code``int` `main() {``      ``// Function Call``    ``cout << numberOfPaths(3, 3);``    ``return` `0;``}` `// This code is contributed by Chandramani Kumar`

Output

`13`

Time Complexity: O(M*N) as two nested loops are executing one from 1 to M and other from 1 to N where M and N are rows and columns respectively.

Space Complexity: O(M*N) as 2D array dp has been created of size M+1 and N+1 where M and N are rows and columns respectively.

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