Add solution #2684

Author: JoshCrozier

README.md

@@ -1986,6 +1986,7 @@
 2680|[Maximum OR](./solutions/2680-maximum-or.js)|Medium|
 2682|[Find the Losers of the Circular Game](./solutions/2682-find-the-losers-of-the-circular-game.js)|Easy|
 2683|[Neighboring Bitwise XOR](./solutions/2683-neighboring-bitwise-xor.js)|Medium|
+2684|[Maximum Number of Moves in a Grid](./solutions/2684-maximum-number-of-moves-in-a-grid.js)|Medium|
 2685|[Count the Number of Complete Components](./solutions/2685-count-the-number-of-complete-components.js)|Medium|
 2693|[Call Function with Custom Context](./solutions/2693-call-function-with-custom-context.js)|Medium|
 2694|[Event Emitter](./solutions/2694-event-emitter.js)|Medium|

solutions/2684-maximum-number-of-moves-in-a-grid.js

@@ -0,0 +1,53 @@
+/**
+ * 2684. Maximum Number of Moves in a Grid
+ * https://leetcode.com/problems/maximum-number-of-moves-in-a-grid/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed m x n matrix grid consisting of positive integers.
+ *
+ * You can start at any cell in the first column of the matrix, and traverse the grid in the
+ * following way:
+ * - From a cell (row, col), you can move to any of the cells: (row - 1, col + 1), (row, col + 1)
+ *   and (row + 1, col + 1) such that the value of the cell you move to, should be strictly bigger
+ *   than the value of the current cell.
+ *
+ * Return the maximum number of moves that you can perform.
+ */
+
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maxMoves = function(grid) {
+  const rows = grid.length;
+  const cols = grid[0].length;
+  const memo = Array.from({ length: rows }, () => new Array(cols).fill(-1));
+
+  let result = 0;
+  for (let row = 0; row < rows; row++) {
+    result = Math.max(result, explore(row, 0));
+  }
+
+  return result;
+
+  function explore(row, col) {
+    if (col === cols - 1) return 0;
+    if (memo[row][col] !== -1) return memo[row][col];
+
+    let maxSteps = 0;
+    const current = grid[row][col];
+
+    if (row > 0 && col + 1 < cols && grid[row - 1][col + 1] > current) {
+      maxSteps = Math.max(maxSteps, 1 + explore(row - 1, col + 1));
+    }
+    if (col + 1 < cols && grid[row][col + 1] > current) {
+      maxSteps = Math.max(maxSteps, 1 + explore(row, col + 1));
+    }
+    if (row + 1 < rows && col + 1 < cols && grid[row + 1][col + 1] > current) {
+      maxSteps = Math.max(maxSteps, 1 + explore(row + 1, col + 1));
+    }
+
+    memo[row][col] = maxSteps;
+    return maxSteps;
+  }
+};