Create 1463. Cherry Pickup II.cpp

In this, I solved leet code  todays problem that Cherry-pickup-2.

Author: harshuCodes-git

cpp/1463. Cherry Pickup II.cpp

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+//   https://leetcode.com/problems/cherry-pickup-ii/
+
+
+// You are given a rows x cols matrix grid representing a field of cherries where grid[i][j] represents the number of cherries that you can collect from the (i, j) cell.
+
+// You have two robots that can collect cherries for you:
+
+// Robot #1 is located at the top-left corner (0, 0), and
+// Robot #2 is located at the top-right corner (0, cols - 1).
+// Return the maximum number of cherries collection using both robots by following the rules below:
+
+// From a cell (i, j), robots can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
+// When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.
+// When both robots stay in the same cell, only one takes the cherries.
+// Both robots cannot move outside of the grid at any moment.
+// Both robots should reach the bottom row in grid.
+
+
+Solution:
+
+#include <vector>
+#include <algorithm>
+
+using namespace std;
+
+class Solution {
+public:
+    int cherryPickup(vector<vector<int>>& grid) {
+        int rows = grid.size();
+        int cols = grid[0].size();
+
+        vector<vector<vector<int>>> dp(rows, vector<vector<int>>(cols, vector<int>(cols, 0)));
+
+        for (int i = rows - 1; i >= 0; --i) {
+            for (int j = 0; j < cols; ++j) {
+                for (int k = 0; k < cols; ++k) {
+                    int cherries = grid[i][j] + (j != k ? grid[i][k] : 0);
+                    if (i == rows - 1) {
+                        dp[i][j][k] = cherries;
+                    } else {
+                        int maxCherries = 0;
+                        for (int dj = -1; dj <= 1; ++dj) {
+                            for (int dk = -1; dk <= 1; ++dk) {
+                                int nj = j + dj;
+                                int nk = k + dk;
+                                if (nj >= 0 && nj < cols && nk >= 0 && nk < cols) {
+                                    maxCherries = max(maxCherries, dp[i + 1][nj][nk]);
+                                }
+                            }
+                        }
+                        dp[i][j][k] = cherries + maxCherries;
+                    }
+                }
+            }
+        }
+        
+        return dp[0][0][cols - 1];
+    }
+};