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 :



title-img




                        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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