LeetCode 3363: Find the Maximum Number of Fruits Collected Solution
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3363. Find the Maximum Number of Fruits Collected
Problem Explanation
Explanation:
To solve this problem, we can use dynamic programming to keep track of the maximum number of fruits collected by each child at each cell in the dungeon. We can then calculate the total fruits collected by all children at the destination cell by summing up the fruits collected by each child.
- Initialize a 3D DP array of size n x n x 3 where dp[i][j][k] represents the maximum number of fruits that can be collected by the k-th child starting from cell (i, j).
- Iterate through the dungeon cells row by row and column by column, updating the DP array based on the rules mentioned in the problem description.
- Finally, return the sum of fruits collected by all children at the destination cell.
Time Complexity: O(n^2) where n is the size of the dungeon grid. Space Complexity: O(n^2) for the DP array.
Solution Code
class Solution {
public int maxFruits(int[][] fruits) {
int n = fruits.length;
int[][][] dp = new int[n][n][3];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < 3; k++) {
dp[i][j][k] = fruits[i][j];
if (i > 0 && j > 0) dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j - 1][k] + fruits[i][j]);
if (i > 0) dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j][k] + fruits[i][j]);
if (j > 0) dp[i][j][k] = Math.max(dp[i][j][k], dp[i][j - 1][k] + fruits[i][j]);
}
}
}
return dp[n - 1][n - 1][0] + dp[n - 1][n - 1][1] + dp[n - 1][n - 1][2] - fruits[n - 1][n - 1];
}
}Try It Yourself
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