LeetCode in Kotlin

2328. Number of Increasing Paths in a Grid

Hard

You are given an m x n integer matrix grid, where you can move from a cell to any adjacent cell in all 4 directions.

Return the number of strictly increasing paths in the grid such that you can start from any cell and end at any cell. Since the answer may be very large, return it modulo 109 + 7.

Two paths are considered different if they do not have exactly the same sequence of visited cells.

Example 1:

Input: grid = [[1,1],[3,4]]

Output: 8

Explanation: The strictly increasing paths are:

• Paths with length 1: [1], [1], [3], [4].

• Paths with length 2: [1 -> 3], [1 -> 4], [3 -> 4].

• Paths with length 3: [1 -> 3 -> 4].

The total number of paths is 4 + 3 + 1 = 8.

Example 2:

Input: grid = [[1],[2]]

Output: 3

Explanation: The strictly increasing paths are:

• Paths with length 1: [1], [2].

• Paths with length 2: [1 -> 2].

The total number of paths is 2 + 1 = 3.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 1000
• 1 <= m * n <= 105
• 1 <= grid[i][j] <= 105

Solution

class Solution {
    private fun help(a: Array<IntArray>, i: Int, j: Int, n: Int, m: Int, dp: Array<IntArray>): Int {
        if (i < 0 || i >= n || j >= m || j < 0) {
            return 0
        }
        if (dp[i][j] != 0) {
            return dp[i][j]
        }
        var res: Long = 0
        if (i < n - 1 && a[i + 1][j] > a[i][j]) {
            res += (1 + help(a, i + 1, j, n, m, dp)).toLong()
        }
        if (i > 0 && a[i - 1][j] > a[i][j]) {
            res += (1 + help(a, i - 1, j, n, m, dp)).toLong()
        }
        if (j > 0 && a[i][j - 1] > a[i][j]) {
            res += (1 + help(a, i, j - 1, n, m, dp)).toLong()
        }
        if (j < m - 1 && a[i][j + 1] > a[i][j]) {
            res += (1 + help(a, i, j + 1, n, m, dp)).toLong()
        }
        dp[i][j] = res.toInt() % 1000000007
        return dp[i][j]
    }

    fun countPaths(grid: Array<IntArray>): Int {
        val n = grid.size
        val m = grid[0].size
        var ans = n.toLong() * m
        val dp = Array(n) { IntArray(m) }
        for (i in 0 until n) {
            for (j in 0 until m) {
                ans += (help(grid, i, j, n, m, dp) % 1000000007).toLong()
            }
        }
        ans %= 1000000007
        return ans.toInt()
    }
}

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