# Count number of ways to reach a given score in a Matrix

Given a N x N matrix mat[][] consisting of non-negative integers, the task is to find the number of ways to reach a given score M starting from the cell (0, 0) and reaching the cell (N – 1, N – 1) by going only down(from (i, j) to (i + 1, j)) or right(from (i, j) to (i, j + 1)). Whenever a cell (i, j) is reached, total score is updated by currentScore + mat[i][j].
Examples:

Input: mat[][] = {{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}, M = 4
Output:
All the paths will result in a score of 5.
Input: mat[][] = {{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}, M = 5
Output:

Approach: This problem can be solved using dynamic programming. First, we need to decide the states of the DP. For every cell (i, j) and a number 0 ? X ? M, we will store the number of ways to reach that cell from the cell (0, 0) with a total score of X. Thus, our solution will use 3-dimensional dynamic programming, two for the coordinates of the cells and one for the required score value.
The required recurrence relation will be,

dp[i][j][M] = dp[i – 1][j][M – mat[i][j]] + dp[i][j-1][M – mat[i][j]]

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;` `#define n 3` `#define MAX 30`   `// To store the states of dp` `int` `dp[n][n][MAX];`   `// To check whether a particular state` `// of dp has been solved` `bool` `v[n][n][MAX];`   `// Function to find the ways` `// using memoization` `int` `findCount(``int` `mat[][n], ``int` `i, ``int` `j, ``int` `m)` `{` `    ``// Base cases` `    ``if` `(i == 0 && j == 0) {` `        ``if` `(m == mat[0][0])` `            ``return` `1;` `        ``else` `            ``return` `0;` `    ``}`   `    ``// If required score becomes negative` `    ``if` `(m < 0)` `        ``return` `0;`   `    ``if` `(i < 0 || j < 0)` `        ``return` `0;`   `    ``// If current state has been reached before` `    ``if` `(v[i][j][m])` `        ``return` `dp[i][j][m];`   `    ``// Set current state to visited` `    ``v[i][j][m] = ``true``;`   `    ``dp[i][j][m] = findCount(mat, i - 1, j, m - mat[i][j])` `                  ``+ findCount(mat, i, j - 1, m - mat[i][j]);` `    ``return` `dp[i][j][m];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `mat[n][n] = { { 1, 1, 1 },` `                      ``{ 1, 1, 1 },` `                      ``{ 1, 1, 1 } };` `    ``int` `m = 5;` `    ``cout << findCount(mat, n - 1, n - 1, m);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `class` `GFG` `{`   `static` `int` `n = ``3``;` `static` `int` `MAX =``30``;`   `// To store the states of dp` `static` `int` `dp[][][] = ``new` `int``[n][n][MAX];`   `// To check whether a particular state` `// of dp has been solved` `static` `boolean` `v[][][] = ``new` `boolean``[n][n][MAX];`   `// Function to find the ways` `// using memoization` `static` `int` `findCount(``int` `mat[][], ``int` `i, ``int` `j, ``int` `m)` `{` `    ``// Base cases` `    ``if` `(i == ``0` `&& j == ``0``) ` `    ``{` `        ``if` `(m == mat[``0``][``0``])` `            ``return` `1``;` `        ``else` `            ``return` `0``;` `    ``}`   `    ``// If required score becomes negative` `    ``if` `(m < ``0``)` `        ``return` `0``;`   `    ``if` `(i < ``0` `|| j < ``0``)` `        ``return` `0``;`   `    ``// If current state has been reached before` `    ``if` `(v[i][j][m])` `        ``return` `dp[i][j][m];`   `    ``// Set current state to visited` `    ``v[i][j][m] = ``true``;`   `    ``dp[i][j][m] = findCount(mat, i - ``1``, j, m - mat[i][j])` `                ``+ findCount(mat, i, j - ``1``, m - mat[i][j]);` `    ``return` `dp[i][j][m];` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `mat[][] = { { ``1``, ``1``, ``1` `},` `                    ``{ ``1``, ``1``, ``1` `},` `                    ``{ ``1``, ``1``, ``1` `} };` `    ``int` `m = ``5``;` `    ``System.out.println(findCount(mat, n - ``1``, n - ``1``, m));` `}` `}`   `/* This code contributed by PrinciRaj1992 */`

## Python3

 `# Python3 implementation of the approach` `n ``=` `3` `MAX` `=` `60`   `# To store the states of dp` `dp ``=` `[[[``0` `for` `i ``in` `range``(``30``)]   ` `          ``for` `i ``in` `range``(``30``)]` `          ``for` `i ``in` `range``(``MAX` `+` `1``)]`   `# To check whether a particular state` `# of dp has been solved` `v ``=` `[[[``0` `for` `i ``in` `range``(``30``)] ` `         ``for` `i ``in` `range``(``30``)]` `         ``for` `i ``in` `range``(``MAX` `+` `1``)]`   `# Function to find the ways` `# using memoization` `def` `findCount(mat, i, j, m):` `    `  `    ``# Base cases` `    ``if` `(i ``=``=` `0` `and` `j ``=``=` `0``):` `        ``if` `(m ``=``=` `mat[``0``][``0``]):` `            ``return` `1` `        ``else``:` `            ``return` `0`   `    ``# If required score becomes negative` `    ``if` `(m < ``0``):` `        ``return` `0`   `    ``if` `(i < ``0` `or` `j < ``0``):` `        ``return` `0`   `    ``# If current state has been reached before` `    ``if` `(v[i][j][m] > ``0``):` `        ``return` `dp[i][j][m]`   `    ``# Set current state to visited` `    ``v[i][j][m] ``=` `True`   `    ``dp[i][j][m] ``=` `(findCount(mat, i ``-` `1``, j, ` `                             ``m ``-` `mat[i][j]) ``+` `                   ``findCount(mat, i, j ``-` `1``, ` `                             ``m ``-` `mat[i][j]))`   `    ``return` `dp[i][j][m]`   `# Driver code` `mat ``=` `[ [ ``1``, ``1``, ``1` `],` `        ``[ ``1``, ``1``, ``1` `],` `        ``[ ``1``, ``1``, ``1` `] ]` `m ``=` `5` `print``(findCount(mat, n ``-` `1``, n ``-` `1``, m))`   `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach ` `using` `System;`   `class` `GFG ` `{ `   `static` `int` `n = 3; ` `static` `int` `MAX = 30; `   `// To store the states of dp ` `static` `int` `[,,]dp = ``new` `int``[n, n, MAX]; `   `// To check whether a particular state ` `// of dp has been solved ` `static` `bool` `[,,]v = ``new` `bool``[n, n, MAX]; `   `// Function to find the ways ` `// using memoization ` `static` `int` `findCount(``int` `[,]mat, ``int` `i,` `                          ``int` `j, ``int` `m) ` `{ ` `    ``// Base cases ` `    ``if` `(i == 0 && j == 0) ` `    ``{ ` `        ``if` `(m == mat[0, 0]) ` `            ``return` `1; ` `        ``else` `            ``return` `0; ` `    ``} `   `    ``// If required score becomes negative ` `    ``if` `(m < 0) ` `        ``return` `0; `   `    ``if` `(i < 0 || j < 0) ` `        ``return` `0; `   `    ``// If current state has been reached before ` `    ``if` `(v[i, j, m]) ` `        ``return` `dp[i, j, m]; `   `    ``// Set current state to visited ` `    ``v[i, j, m] = ``true``; `   `    ``dp[i, j, m] = findCount(mat, i - 1, j, m - mat[i, j]) + ` `                  ``findCount(mat, i, j - 1, m - mat[i, j]); ` `    ``return` `dp[i, j, m]; ` `} `   `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[,]mat = {{ 1, 1, 1 }, ` `                  ``{ 1, 1, 1 }, ` `                  ``{ 1, 1, 1 }}; ` `    ``int` `m = 5; ` `    ``Console.WriteLine(findCount(mat, n - 1, n - 1, m)); ` `}` `} `   `// This code is contributed by Ryuga `

## PHP

 `

## Javascript

 ``

Output:

`6`

Time complexity: O(N * N * M)

Auxiliary Space: O(N * N * MAX)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next