**Collecting Coins - Amazon Top Interview Questions**

### Problem Statement :

You are given a two-dimensional integer matrix where each cell represents number of coins in that cell. Assuming we start at matrix[0][0], and can only move right or down, find the maximum number of coins you can collect by the bottom right corner. Constraints n, m ≤ 100 where n and m are the number of rows and columns in matrix. Example 1 Input matrix = [ [0, 3, 1, 1], [2, 0, 0, 4] ] Output 9 Explanation We take the following path: [0, 3, 1, 1, 4] Example 2 Input matrix = [ [0, 3, 1, 1], [2, 0, 0, 4], [1, 5, 3, 1] ] Output 12 Explanation We take the following path: [0, 2, 1, 5, 3, 1] Example 3 Input matrix = [ [0, 2, 1], [2, 5, 0], [4, 1, 3] ] Output 11 Explanation We take the following path: [0, 2, 5, 1, 3]

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
vector<vector<int>> dp(int(matrix.size()), vector<int>(int(matrix[0].size()), 0));
dp[0][0] = matrix[0][0];
for (int j = 0; j < int(matrix.size()); ++j) {
for (int i = 0; i < int(matrix[0].size()); ++i) {
if (i - 1 >= 0) dp[j][i] = max(dp[j][i - 1] + matrix[j][i], dp[j][i]);
if (j - 1 >= 0) dp[j][i] = max(dp[j - 1][i] + matrix[j][i], dp[j][i]);
}
}
return dp[int(matrix.size()) - 1][int(matrix[0].size()) - 1];
}
```

` ````
Solution in Python :
class Solution:
def solve(self, matrix):
for i in range(len(matrix)):
for j in range(len(matrix[0])):
fromLeft = matrix[i][j]
fromAbove = matrix[i][j]
if j > 0:
fromLeft = matrix[i][j] + matrix[i][j - 1]
if i > 0:
fromAbove = matrix[i][j] + matrix[i - 1][j]
matrix[i][j] = max(fromLeft, fromAbove)
return matrix[-1][-1]
```

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