Description
You are given a twodimensional 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
wheren
andm
are the number of rows and columns inmatrix
.
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]
Intuition
we need to calculate the sum of coins in every cells, which equals max(up, left) + coins
Implementation
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
int[][] dp = new int[m + 1][n + 1];
for (int row = 1; row <= m; row++) {
for (int col = 1; col <= n; col++) {
dp[row][col] =
Math.max(dp[row  1][col], dp[row][col  1]) + matrix[row  1][col  1];
}
}
return dp[m][n];
}
}
Time Complexity
 Time: O(m*n)
 Space: O(m*n)
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