Add solution #741

Author: JoshCrozier

README.md

@@ -562,6 +562,7 @@
 738|[Monotone Increasing Digits](./0738-monotone-increasing-digits.js)|Medium|
 739|[Daily Temperatures](./0739-daily-temperatures.js)|Medium|
 740|[Delete and Earn](./0740-delete-and-earn.js)|Medium|
+741|[Cherry Pickup](./0741-cherry-pickup.js)|Hard|
 743|[Network Delay Time](./0743-network-delay-time.js)|Medium|
 744|[Find Smallest Letter Greater Than Target](./0744-find-smallest-letter-greater-than-target.js)|Easy|
 745|[Prefix and Suffix Search](./0745-prefix-and-suffix-search.js)|Hard|

solutions/0741-cherry-pickup.js

@@ -0,0 +1,63 @@
+/**
+ * 741. Cherry Pickup
+ * https://leetcode.com/problems/cherry-pickup/
+ * Difficulty: Hard
+ *
+ * You are given an n x n grid representing a field of cherries, each cell is one of three
+ * possible integers.
+ * - 0 means the cell is empty, so you can pass through,
+ * - 1 means the cell contains a cherry that you can pick up and pass through, or
+ * - -1 means the cell contains a thorn that blocks your way.
+ *
+ * Return the maximum number of cherries you can collect by following the rules below:
+ * - Starting at the position (0, 0) and reaching (n - 1, n - 1) by moving right or down through
+ *   valid path cells (cells with value 0 or 1).
+ * - After reaching (n - 1, n - 1), returning to (0, 0) by moving left or up through valid path
+ *   cells.
+ * - When passing through a path cell containing a cherry, you pick it up, and the cell becomes
+ *   an empty cell 0.
+ * - If there is no valid path between (0, 0) and (n - 1, n - 1), then no cherries can be collected.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var cherryPickup = function(grid) {
+  const n = grid.length;
+  const dp = Array.from({ length: n }, () =>
+    Array.from({ length: n }, () =>
+      new Array(2 * n - 1).fill(-1))
+  );
+
+  if (grid[0][0] === -1) return 0;
+  dp[0][0][0] = grid[0][0];
+
+  for (let t = 1; t <= 2 * n - 2; t++) {
+    for (let x1 = Math.max(0, t - n + 1); x1 <= Math.min(n - 1, t); x1++) {
+      for (let x2 = Math.max(0, t - n + 1); x2 <= Math.min(n - 1, t); x2++) {
+        const y1 = t - x1;
+        const y2 = t - x2;
+
+        if (grid[x1][y1] === -1 || grid[x2][y2] === -1) continue;
+
+        const cherries = grid[x1][y1] + (x1 === x2 ? 0 : grid[x2][y2]);
+        let maxPrev = -1;
+
+        for (const px1 of [x1 - 1, x1]) {
+          for (const px2 of [x2 - 1, x2]) {
+            if (px1 >= 0 && px2 >= 0 && dp[px1][px2][t - 1] >= 0) {
+              maxPrev = Math.max(maxPrev, dp[px1][px2][t - 1]);
+            }
+          }
+        }
+
+        if (maxPrev >= 0) {
+          dp[x1][x2][t] = maxPrev + cherries;
+        }
+      }
+    }
+  }
+
+  return dp[n - 1][n - 1][2 * n - 2] < 0 ? 0 : dp[n - 1][n - 1][2 * n - 2];
+};