Suppose, we have a 2D matrix in which the cells represent the number of coins in it. There are our two friends to collect coins, and they are placed at the top left corner and at the top right corner at the start. They follow these rules:
From cell (i, j), a coin-collector can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
Upon reaching a cell they collect all the coins available making the cell empty.
The collectors may choose to stay at one cell, but the coins in any cell can be collected only once.
We have to find the maximum number of coins that can be collected.
So, if the input is like
| 0 | 4 | 1 | 0 |
| 3 | 1 | 4 | 0 |
| 2 | 5 | 1 | 1 |
| 3 | 0 | 0 | 0 |
then the output will be 17.
To solve this, we will follow these steps −
A := the input matrix
R := row count of A
C := column count of A
Define a function dp() . This will take r, c1, c2
if r is same as R, then
return 0
ans := A[r, c1] +(if c1 is not equal to c2, then 1 else 0) * A[r, c2]
base := ans
for each nc1 in [c1 − 1, c1, c1 + 1], do
for each nc2 in [c2 − 1, c2, c2 + 1], do
if 0 <= nc1 < C and 0 <= nc2 < C, then
ans := maximum of ans and (base + dp(r + 1, nc1, nc2))
return ans
return dp(0, 0, C − 1)
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, A): R, C = len(A), len(A[0]) def dp(r, c1, c2): if r == R: return 0 ans = base = A[r][c1] + (c1 != c2) * A[r][c2] for nc1 in [c1 − 1, c1, c1 + 1]: for nc2 in [c2 − 1, c2, c2 + 1]: if 0 <= nc1 < C and 0 <= nc2 < C: ans = max(ans, base + dp(r + 1, nc1, nc2)) return ans return dp(0, 0, C − 1) ob = Solution() print(ob.solve([ [0, 4, 1, 0], [3, 1, 4, 0], [2, 5, 1, 1], [3, 0, 0, 0] ]))
Input
[ [0, 4, 1, 0], [3, 1, 4, 0], [2, 5, 1, 1], [3, 0, 0, 0] ]
Output
17