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

• Difficulty Level : Medium
• Last Updated : 07 Jul, 2021

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)``            ``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

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## Javascript

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Output:
`6`

Time complexity: O(N * N * M)

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