**Collecting Disappearing Coins - Amazon Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers matrix where each matrix[r][c] represents the number of coins in that cell. When you pick up coins on matrix[r][c], all the coins on row r - 1 and r + 1 disappear, as well as the coins at the two cells matrix[r][c + 1] and matrix[r][c - 1]. Return the maximum number of coins that you can collect. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix Example 1 Input matrix = [ [1, 7, 6, 5], [9, 9, 3, 1], [4, 8, 1, 2] ] Output 22 Explanation We can pick cells with the coins 7, 5, and 8 and 2.

### Solution :

` ````
Solution in C++ :
int best(vector<int>& v) {
// return the maximum sum of elements given that you can't choose adjacent elements
int n = v.size();
// dp[i] is the maximum sum you can get considering exactly the first i elements
vector<int> dp(n + 1);
for (int i = 0; i < n; i++) {
// don't choose element i
dp[i + 1] = max(dp[i + 1], dp[i]);
// choose element i
dp[i + 1] = max(dp[i + 1], v[i] + (i == 0 ? 0 : dp[i - 1]));
}
return dp[n];
}
int solve(vector<vector<int>>& matrix) {
vector<int> rows;
for (auto& row : matrix) {
rows.push_back(best(row));
}
return best(rows);
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
if (matrix == null || matrix.length == 0)
return 0;
int[] arr = new int[matrix.length];
for (int i = 0; i < matrix.length; i++) arr[i] = getMaximumNonAdjacentSum(matrix[i]);
return getMaximumNonAdjacentSum(arr);
}
private int getMaximumNonAdjacentSum(int[] arr) {
if (arr.length == 0)
return 0;
if (arr.length == 1)
return arr[0];
if (arr.length == 2)
return Math.max(arr[0], arr[1]);
int[] dp = new int[arr.length];
dp[0] = arr[0];
dp[1] = Math.max(arr[0], arr[1]);
for (int i = 2; i < arr.length; i++) dp[i] = Math.max(dp[i - 1], arr[i] + dp[i - 2]);
return dp[arr.length - 1];
}
}
```

` ````
Solution in Python :
class Solution:
def best(self, l):
canadd = 0
cannotadd = 0
for elem in l:
if isinstance(elem, list):
elem = self.best(elem)
canadd, cannotadd = cannotadd, max(cannotadd, canadd + elem)
return cannotadd
def solve(self, matrix):
return self.best(matrix)
```

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