Number of paths with exactly k coins

Given a matrix where every cell has some number of coins. Count number of ways to reach bottom right from top left with exactly k coins. We can move to (i+1, j) and (i, j+1) from a cell (i, j).

Example:

Input:  k = 12
        mat[][] = { {1, 2, 3},
                    {4, 6, 5},
                    {3, 2, 1}
                  };
Output:  2
There are two paths with 12 coins
1 -> 2 -> 6 -> 2 -> 1
1 -> 2 -> 3 -> 5 -> 1

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The above problem can be recursively defined as below:

pathCount(m, n, k):   Number of paths to reach mat[m][n] from mat[0][0] 
                      with exactly k coins

If (m == 0 and n == 0)
   return 1 if mat[0][0] == k else return 0
Else:
    pathCount(m, n, k) = pathCount(m-1, n, k - mat[m][n]) + 
                         pathCount(m, n-1, k - mat[m][n])

Below is the implementation of above recursive algorithm.

C++

    
  
// A Naive Recursive C++ program  
// to count paths with exactly 
// 'k' coins 
#include <bits/stdc++.h> 
#define R 3 
#define C 3 
using namespace std; 
  
// Recursive function to count paths with sum k from 
// (0, 0) to (m, n) 
int pathCountRec(int mat[][C], int m, int n, int k) 
{ 
    // Base cases 
    if (m < 0 || n < 0) return 0; 
    if (m==0 && n==0) return (k == mat[m][n]); 
  
    // (m, n) can be reached either through (m-1, n) or 
    // through (m, n-1) 
    return pathCountRec(mat, m-1, n, k-mat[m][n]) + 
           pathCountRec(mat, m, n-1, k-mat[m][n]); 
} 
  
// A wrapper over pathCountRec() 
int pathCount(int mat[][C], int k) 
{ 
    return pathCountRec(mat, R-1, C-1, k); 
} 
  
// Driver program 
int main() 
{ 
    int k = 12; 
    int mat[R][C] = { {1, 2, 3}, 
                      {4, 6, 5}, 
                      {3, 2, 1} 
                  }; 
    cout << pathCount(mat, k); 
    return 0; 
}

Java

// A Naive Recursive Java program to  
// count paths with exactly 'k' coins   
  
class GFG { 
  
    static final int R = 3; 
    static final int C = 3; 
  
// Recursive function to count paths with sum k from  
// (0, 0) to (m, n)  
    static int pathCountRec(int mat[][], int m, int n, int k) { 
        // Base cases  
        if (m < 0 || n < 0) { 
            return 0; 
        } 
        if (m == 0 && n == 0 && (k == mat[m][n])) { 
            return 1; 
        } 
  
        // (m, n) can be reached either through (m-1, n) or  
        // through (m, n-1)  
        return pathCountRec(mat, m - 1, n, k - mat[m][n]) 
                + pathCountRec(mat, m, n - 1, k - mat[m][n]); 
    } 
  
// A wrapper over pathCountRec()  
    static int pathCount(int mat[][], int k) { 
        return pathCountRec(mat, R - 1, C - 1, k); 
    } 
  
    // Driver code  
    public static void main(String[] args) { 
        int k = 12; 
        int mat[][] = {{1, 2, 3}, 
        {4, 6, 5}, 
        {3, 2, 1} 
        }; 
        System.out.println(pathCount(mat, k)); 
    } 
} 
  
/* This Java code is contributed by Rajput-Ji*/

Python3

# A Naive Recursive Python program to 
# count paths with exactly 'k' coins 
  
R = 3
C = 3
  
# Recursive function to count paths 
# with sum k from (0, 0) to (m, n) 
def pathCountRec(mat, m, n, k): 
  
    # Base cases 
    if m < 0 or n < 0: 
        return 0
    elif m == 0 and n == 0: 
        return k == mat[m][n] 
  
    # #(m, n) can be reached either 
    # through (m-1, n) or through 
    # (m, n-1) 
    return (pathCountRec(mat, m-1, n, k-mat[m][n])  
         + pathCountRec(mat, m, n-1, k-mat[m][n])) 
  
# A wrapper over pathCountRec() 
def pathCount(mat, k): 
    return pathCountRec(mat, R-1, C-1, k) 
  
# Driver Program 
k = 12
mat = [[1, 2, 3], 
       [4, 6, 5], 
       [3, 2, 1]] 
  
print(pathCount(mat, k)) 
  
# This code is contributed by Shrikant13. 

C#

using System; 
  
// A Naive Recursive c# program to  
// count paths with exactly 'k' coins  
  
public class GFG 
{ 
  
    public const int R = 3; 
    public const int C = 3; 
  
// Recursive function to count paths with sum k from  
// (0, 0) to (m, n)  
    public static int pathCountRec(int[][] mat, int m, int n, int k) 
    { 
        // Base cases  
        if (m < 0 || n < 0) 
        { 
            return 0; 
        } 
        if (m == 0 && n == 0 && (k == mat[m][n])) 
        { 
            return 1; 
        } 
  
        // (m, n) can be reached either through (m-1, n) or  
        // through (m, n-1)  
        return pathCountRec(mat, m - 1, n, k - mat[m][n])  
                + pathCountRec(mat, m, n - 1, k - mat[m][n]); 
    } 
  
// A wrapper over pathCountRec()  
    public static int pathCount(int[][] mat, int k) 
    { 
        return pathCountRec(mat, R - 1, C - 1, k); 
    } 
  
    // Driver code  
    public static void Main(string[] args) 
    { 
        int k = 12; 
        int[][] mat = new int[][] 
        { 
            new int[] {1, 2, 3}, 
            new int[] {4, 6, 5}, 
            new int[] {3, 2, 1} 
        }; 
        Console.WriteLine(pathCount(mat, k)); 
    } 
} 
  
// This code is contributed by Shrikant13 

PHP

<?php 
// A Naive Recursive PHP program to 
// count paths with exactly 'k' coins 
  
$R = 3; 
$C = 3; 
  
// Recursive function to count paths 
// with sum k from (0, 0) to (m, n) 
function pathCountRec( $mat, $m, $n, $k) 
{ 
      
    // Base cases 
    if ($m < 0 or $n < 0)  
        return 0; 
    if ($m == 0 and $n == 0)  
        return ($k == $mat[$m][$n]); 
  
    // (m, n) can be reached either  
    // through (m-1, n) or through  
    // (m, n-1) 
    return pathCountRec($mat, $m - 1, 
                $n, $k - $mat[$m][$n]) 
              + pathCountRec($mat, $m, 
           $n - 1, $k - $mat[$m][$n]); 
} 
  
// A wrapper over pathCountRec() 
function pathCount($mat, $k) 
{ 
    global $R, $C; 
    return pathCountRec($mat, $R-1,  
                            $C-1, $k); 
} 
  
// Driver program 
  
    $k = 12; 
    $mat = array(array(1, 2, 3), 
                 array(4, 6, 5), 
                 array(3, 2, 1) ); 
                   
    echo pathCount($mat, $k); 
  
// This code is contributed by anuj_67. 
?> 

Output:

The time complexity of above solution recursive is exponential. We can solve this problem in Pseudo Polynomial Time (time complexity is dependent on numeric value of input) using Dynamic Programming. The idea is to use a 3 dimensional table dp[m][n][k] where m is row number, n is column number and k is number of coins. Below is Dynamic Programming based the implementation.

C++

// A Dynamic Programming based C++ program to count paths with 
// exactly 'k' coins 
#include <bits/stdc++.h> 
#define R 3 
#define C 3 
#define MAX_K 1000 
using namespace std; 
  
int dp[R][C][MAX_K]; 
  
int pathCountDPRecDP(int mat[][C], int m, int n, int k) 
{ 
    // Base cases 
    if (m < 0 || n < 0) return 0; 
    if (m==0 && n==0) return (k == mat[m][n]); 
  
    // If this subproblem is already solved 
    if (dp[m][n][k] != -1) return dp[m][n][k]; 
  
    // (m, n) can be reached either through (m-1, n) or 
    // through (m, n-1) 
    dp[m][n][k] = pathCountDPRecDP(mat, m-1, n, k-mat[m][n]) + 
                  pathCountDPRecDP(mat, m, n-1, k-mat[m][n]); 
  
    return dp[m][n][k]; 
} 
  
// This function mainly initializes dp[][][] and calls 
// pathCountDPRecDP() 
int pathCountDP(int mat[][C], int k) 
{ 
    memset(dp, -1, sizeof dp); 
    return pathCountDPRecDP(mat, R-1, C-1, k); 
} 
  
// Driver Program to test above functions 
int main() 
{ 
    int k = 12; 
    int mat[R][C] = { {1, 2, 3}, 
                      {4, 6, 5}, 
                      {3, 2, 1} 
                  }; 
    cout << pathCountDP(mat, k); 
    return 0; 
} 

Java

// A Dynamic Programming based JAVA program to count paths with 
// exactly 'k' coins 
  
  
class GFG 
{ 
      
    static final int R = 3; 
    static final int C = 3; 
    static final int MAX_K = 100; 
      
    static int [][][]dp=new int[R][C][MAX_K]; 
      
    static int pathCountDPRecDP(int [][]mat, int m, int n, int k) 
    { 
        // Base cases 
        if (m < 0 || n < 0) return 0; 
        if (m==0 && n==0) return (k == mat[m][n] ? 1 : 0); 
      
        // If this subproblem is already solved 
        if (dp[m][n][k] != -1) return dp[m][n][k]; 
      
        // (m, n) can be reached either through (m-1, n) or 
        // through (m, n-1) 
        dp[m][n][k] = pathCountDPRecDP(mat, m-1, n, k-mat[m][n]) + 
                    pathCountDPRecDP(mat, m, n-1, k-mat[m][n]); 
      
        return dp[m][n][k]; 
    } 
      
    // This function mainly initializes dp[][][] and calls 
    // pathCountDPRecDP() 
    static int pathCountDP(int [][]mat, int k) 
    { 
        for(int i=0;i<R;i++) 
            for(int j=0;j<C;j++) 
                for(int l=0;l<MAX_K;l++) 
                  dp[i][j][l]=-1; 
                    
        return pathCountDPRecDP(mat, R-1, C-1, k); 
    } 
      
    // Driver Program to test above functions 
    public static void main(String []args) 
    { 
        int k = 12; 
        int[][] mat = new int[][] 
        { 
            new int[] {1, 2, 3}, 
            new int[] {4, 6, 5}, 
            new int[] {3, 2, 1} 
        }; 
                      
        System.out.println(pathCountDP(mat, k)); 
          
    } 
  
} 
  
// This code is contributed by ihritik 

Python3

# A Dynamic Programming based Python3 program to 
# count paths with exactly 'k' coins 
R = 3
C = 3
MAX_K = 1000
  
def pathCountDPRecDP(mat, m, n, k): 
  
    # Base cases 
    if m < 0 or n < 0: 
        return 0
    elif m == 0 and n == 0: 
        return k == mat[m][n] 
      
    # If this subproblem is already solved 
    if (dp[m][n][k] != -1): 
        return dp[m][n][k] 
  
    # #(m, n) can be reached either 
    # through (m-1, n) or through 
    # (m, n-1) 
    dp[m][n][k] = (pathCountDPRecDP(mat, m - 1, n, k - mat[m][n]) + 
                   pathCountDPRecDP(mat, m, n - 1, k - mat[m][n])) 
      
    return dp[m][n][k] 
  
# A wrapper over pathCountDPRecDP() 
def pathCountDP(mat, k): 
    return pathCountDPRecDP(mat, R - 1, C - 1, k) 
  
# Driver Code 
k = 12
  
# Initialising dp[][][] 
dp = [[ [-1 for col in range(MAX_K)]  
            for col in range(C)]  
            for row in range(R)] 
  
mat = [[1, 2, 3], 
       [4, 6, 5], 
       [3, 2, 1]] 
  
print(pathCountDP(mat, k)) 
  
# This code is contributed by ashutosh450 

Output:

Time complexity of this solution is O(m*n*k).

Thanks to Gaurav Ahirwar for suggesting above solution.

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