Skip to content
Related Articles

Related Articles

Improve Article

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

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

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 <iostream>
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




<?php
// PHP implementation of the approach
$n = 3;
$MAX = 30;
 
// To store the states of dp
$dp = array($n, $n, $MAX);
 
// To check whether a particular state
// of dp has been solved
$v = array($n, $n, $MAX);
 
// Function to find the ways
// using memoization
function findCount($mat, $i, $j, $m)
{
    // Base cases
    if ($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
$mat = array(array(1, 1, 1 ),
             array(1, 1, 1 ),
             array(1, 1, 1 ));
$m = 5;
echo(findCount($mat, $n - 1, $n - 1, $m));
 
// This code contributed by Code_Mech

Javascript




<script>
 
// Javascript implementation of the approach
var n = 3;
var MAX = 30;
 
// To store the states of dp
var dp = Array(n);
for(var i =0; i<n; i++)
{
    dp[i] = Array(n);
    for(var j=0; j<n; j++)
    {
        dp[i][j]=Array(MAX);
    }
}
 
// To check whether a particular state
// of dp has been solved
var v = Array(n);
for(var i =0; i<n; i++)
{
    v[i] = Array(n);
    for(var j=0; j<n; j++)
    {
        v[i][j]=Array(MAX);
    }
}
 
// Function to find the ways
// using memoization
function findCount(mat, i, j, 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
var mat = [ [ 1, 1, 1 ],
                  [ 1, 1, 1 ],
                  [ 1, 1, 1 ] ];
var m = 5;
document.write( findCount(mat, n - 1, n - 1, m));
 
</script>
Output: 
6

 

Time complexity: O(N * N * M)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :