Program to find maximum amount of coin we can collect from a given matrix in Python

Suppose we have a 2D matrix where matrix [r, c] represents the number of coins in that cell. We can start from any position and want to collect coins by moving any of the four directions (up, down, left and right, not diagonally). When we move to a cell the coins are collected and the value of that cell becomes 0. We cannot visit a cell with 0 coins, we have to find the maximum amount of coins we can collect.

So, if the input is like

then the output will be 18, as we can take the path 2 −> 3 −> 6 −> 4 −> 3

To solve this, we will follow these steps −

• if matrix is empty, then

  • return 0

• n := row count of matrix, m := column count of mat

• x := a list like [−1, 1, 0, 0], y := a list like [0, 0, −1, 1]

• Define a function util() . This will take a, b

• ret := 0

• for k in range 0 to 3, do

  • (t1, t2) := (x[k] + a, y[k] + b)

  • if (t1, t2) is valid cell, then

    • t := mat[t1, t2], mat[t1, t2] := 0

    • ret := maximum of ret and (util(t1, t2) + t)

    • mat[t1, t2] := t

• return ret

• From the main method do the following −

• res := 0

• for i in range 0 to n − 1, do

  • for j in range 0 to m − 1, do

    • if mat[i, j] is non−zero, then

      • t := mat[i, j], mat[i, j] := 0

      • res := maximum of res and (util(i, j) + temp)

• return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, mat):
      if not mat:
         return 0
      n, m = len(mat), len(mat[0])
      x, y = [−1, 1, 0, 0], [0, 0, −1, 1]
      def ok(a, b):
         return 0 <= a < n and 0 <= b < m and mat[a][b]
      def util(a, b):
         ret = 0
         for k in range(4):
            t1, t2 = x[k] + a, y[k] + b
            if ok(t1, t2):
               t, mat[t1][t2] = mat[t1][t2], 0
               ret = max(ret, util(t1, t2) + t)
               mat[t1][t2] = t
            return ret
         res = 0
         for i in range(n):
            for j in range(m):
               if mat[i][j]:
                  temp, mat[i][j] = mat[i][j], 0
                  res = max(res, util(i, j) + temp)
         return res
ob = Solution()
matrix = [
   [2, 4, 3],
   [3, 6, 0],
   [2, 0, 12]
]
print(ob.solve(matrix))

Input

[
[2, 4, 3],
[3, 6, 0],
[2, 0, 12] ]

Output

18