LeetCode 3466: Maximum Coin Collection Solution

Master LeetCode problem 3466 (Maximum Coin Collection ), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3466. Maximum Coin Collection

Medium

Array Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with a 2D array to keep track of the maximum coins collected up to a certain point in the grid. The idea is to start from the top-left corner of the grid and iterate through each cell. At each cell, we calculate the maximum coins that can be collected by considering the maximum coins collected from the top and left cells, and add the current cell's value to it.

Let's define a 2D array dp where dp[i][j] represents the maximum coins collected up to cell (i, j). We can then update this array iteratively by considering the maximum coins from the top and left cells.

The time complexity of this approach is O(mn) where m is the number of rows and n is the number of columns in the grid. The space complexity is also O(mn) for the 2D array to store the maximum coins.

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Solution Code

class Solution {
    public int maxCoins(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        int[][] dp = new int[m][n];
        
        dp[0][0] = grid[0][0];
        
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i - 1][0] + grid[i][0];
        }
        
        for (int j = 1; j < n; j++) {
            dp[0][j] = dp[0][j - 1] + grid[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
            }
        }
        
        return dp[m - 1][n - 1];
    }
}

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