feat: solve NO.2684

Author: BaffinLee

2601-2700/2684. Maximum Number of Moves in a Grid.md

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+# 2684. Maximum Number of Moves in a Grid
+
+- Difficulty: Medium.
+- Related Topics: Array, Dynamic Programming, Matrix.
+- Similar Questions: .
+
+## Problem
+
+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.**
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2023/04/11/yetgriddrawio-10.png)
+
+```
+Input: grid = [[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]]
+Output: 3
+Explanation: We can start at the cell (0, 0) and make the following moves:
+- (0, 0) -> (0, 1).
+- (0, 1) -> (1, 2).
+- (1, 2) -> (2, 3).
+It can be shown that it is the maximum number of moves that can be made.
+```
+
+Example 2:
+
+```
+
+Input: grid = [[3,2,4],[2,1,9],[1,1,7]]
+Output: 0
+Explanation: Starting from any cell in the first column we cannot perform any moves.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `m == grid.length`
+	
+- `n == grid[i].length`
+	
+- `2 <= m, n <= 1000`
+	
+- `4 <= m * n <= 105`
+	
+- `1 <= grid[i][j] <= 106`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var maxMoves = function(grid) {
+    var dp = Array(grid.length).fill(0).map(() => Array(grid[0].length));
+    var max = 0;
+    for (var i = 0; i < grid.length; i++) {
+        max = Math.max(max, helper(grid, i, 0, dp));
+    }
+    return max;
+};
+
+var helper = function(grid, i, j, dp) {
+    if (dp[i][j] !== undefined) return dp[i][j];
+    var max = 0;
+    if (j < grid[0].length - 1) {
+        if (i > 0 && grid[i - 1][j + 1] > grid[i][j]) {
+            max = Math.max(max, 1 + helper(grid, i - 1, j + 1, dp));
+        }
+        if (grid[i][j + 1] > grid[i][j]) {
+            max = Math.max(max, 1 + helper(grid, i, j + 1, dp));
+        }
+        if (i < grid.length - 1 && grid[i + 1][j + 1] > grid[i][j]) {
+            max = Math.max(max, 1 + helper(grid, i + 1, j + 1, dp));
+        }
+    }
+    dp[i][j] = max;
+    return max;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(m * n).
+* Space complexity : O(m * n).