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# Longest Increasing Path in a Matrix

Given an integer matrix, find the length of the longest increasing path.

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

**Example 1:**

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

**Example 2:**

```
Input: nums = 
[
  [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.
```

```java
class Solution {
    int directions[][] = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };

    public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0)
            return 0;
        int dp[][] = new int[matrix.length][matrix[0].length];
        int max = 0;
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[i].length; j++) {
                if (dp[i][j] == 0) {
                    dp[i][j] = maxPath(matrix, dp, i, j);
                }
                max = Math.max(dp[i][j], max);
            }
        }
        return max;
    }

    public int maxPath(int[][] matrix, int dp[][], int x, int y) {
        if (dp[x][y] != 0)
            return dp[x][y];
        dp[x][y] = 1;// Base case
        for (int i = 0; i < 4; i++) {
            int newX = x + directions[i][0], newY = y + directions[i][1];
            if (newX >= 0 && newX < matrix.length && newY >= 0 && newY < matrix[0].length
                    && matrix[x][y] < matrix[newX][newY])
                dp[x][y] = Math.max(dp[x][y], maxPath(matrix, dp, newX, newY) + 1);
        }
        return dp[x][y];
    }
}
```


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