You are given a 0-indexed m x n integer matrix grid. Your initial position is at the top-left cell (0, 0).
Starting from the cell (i, j), you can move to one of the following cells:
(i, k) with j < k <= grid[i][j] + j (rightward movement), or(k, j) with i < k <= grid[i][j] + i (downward movement).Return the minimum number of cells you need to visit to reach the bottom-right cell (m - 1, n - 1). If there is no valid path, return -1.
\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/01/25/ex1.png\")
\n\nInput: grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]\nOutput: 4\nExplanation: The image above shows one of the paths that visits exactly 4 cells.\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/01/25/ex2.png\")
\n\nInput: grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]\nOutput: 3\nExplanation: The image above shows one of the paths that visits exactly 3 cells.\n\n\n
Example 3:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/01/26/ex3.png\")
\n\nInput: grid = [[2,1,0],[1,0,0]]\nOutput: -1\nExplanation: It can be proven that no path exists.\n\n\n
\n
Constraints:
\n\nm == grid.lengthn == grid[i].length1 <= m, n <= 1051 <= m * n <= 1050 <= grid[i][j] < m * ngrid[m - 1][n - 1] == 0Compiled with clang 11 using the latest C++ 17 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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