# Longest Increasing Path in a Matrix

Updated: Apr 10, 2021

Given an m x n integers matrix, return *the length of the longest increasing path in *matrix.

From each cell, you can either move in four directions: left, right, up, or down. You **may not** move **diagonally** or move **outside the boundary** (i.e., wrap-around is not allowed).

**Example 1:**

**Input****:** matrix = [[9,9,4],[6,6,8],[2,1,1]]
**Output****:** 4
**Explanation****:** The longest increasing path is [1, 2, 6, 9].

**Example 2:**

**Input****:** matrix = [[3,4,5],[3,2,6],[2,2,1]]
**Output****:** 4
**Explanation****:**** **The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

**Example 3:**

**Input****:** matrix = [[1]]
**Output****:** 1

**Constraints:**

m == matrix.length

n == matrix[i].length

1 <= m, n <= 200

0 <= matrix[i][j] <= 231 - 1

**Solution:**

```
public class Solution {
public static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public int longestIncreasingPath(int[][] matrix) {
if(matrix.length == 0) return 0;
int m = matrix.length, n = matrix[0].length;
int[][] dp = new int[m][n];
int max = 1;
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
int len = dfs(matrix, i, j, m, n,dp);
max = Math.max(max, len);
}
}
return max;
}
public int dfs(int[][] matrix, int i, int j, int m, int n,int[][] dp)
{
if(dp[i][j]!=0) return dp[i][j];
int max = 1;
for(int[] d: dirs)
{
int x = i + d[0];
int y = j + d[1];
if(x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j]) continue;
int len = 1 + dfs(matrix, x, y, m, n,dp);
max = Math.max(max, len);
}
dp[i][j] = max;
return max;
}
}
```