Program to find number of coins we can pick from topleft to bottom-right cell and return in Python

PythonServer Side ProgrammingProgramming

Suppose we have a 2D matrix with 3 possible values −

  • 0 for an empty cell.

  • 1 for a coin.

  • −1 for a wall.

We have to find the maximum number of coins we can take by starting from the top−left cell and reaching the bottom−right cell by moving only right or down direction. Then come back to the top−left cell by only moving up or left direction. When we pick up a coin, the cell value becomes 0. If we cannot reach the bottom−right cell, then return 0.

So, if the input is like

011
111
−111
011

then the output will be 8.

To solve this, we will follow these steps −

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

  • Define a function util() . This will take i, j, k, l

  • if i and j are not in range of mat or mat[i, j] is same as −1, then

    • return −inf

  • if k and l are not in range of mat or mat[k, l] is same as −1, then

    • return −inf

  • if i, j, k and l all are 0, then

    • return mat[0, 0]

  • best := −inf

  • for each pair (dx1, dy1) in [(−1, 0) ,(0, −1) ], do

    • for each pair (dx2, dy2) in [(−1, 0) ,(0, −1) ], do

      • best := maximum of best and util(i + dy1, j + dx1, k + dy2, l + dx2)

  • return mat[i, j] +(1 when i is not same as k, otherwise 0) * mat[k, l] + best

  • From the main method do the following −

  • return maximum of 0 and util(n − 1, m − 1, n − 1, m − 1)

Let us see the following implementation to get better understanding −

Example

 Live Demo

class Solution:
   def solve(self, mat):
      n, m = len(mat), len(mat[0])
      def util(i, j, k, l):
         if not (0 <= i < n and 0 <= j < m) or mat[i][j] == −1:
            return −1e9
         if not (0 <= k < n and 0 <= l < m) or mat[k][l] == −1:
            return −1e9
         if i == 0 and j == 0 and k == 0 and l == 0:
            return mat[0][0]
         best = −1e9
         for dx1, dy1 in [(−1, 0), (0, −1)]:
            for dx2, dy2 in [(−1, 0), (0, −1)]:
               best = max(best, util(i + dy1, j + dx1, k + dy2, l + dx2))
         return mat[i][j] + (i != k) * mat[k][l] + best
      return max(0, util(n − 1, m − 1, n − 1, m − 1))
ob = Solution()
matrix = [
   [0, 1, 1],
   [1, 1, 1],
   [1, −1, 1],
   [0, 1, 1]
]
print(ob.solve(matrix))

Input

[
   [0, 1, 1],
   [1, 1, 1],
   [1, −1, 1],
   [0, 1, 1]
]

Output

8
raja
Published on 26-Dec-2020 10:47:41
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