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:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]

Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

int dx[] = { 1 , -1, 0 , 0  };
 
int dy[] = { 0 , 0 , 1 , -1 };
class Solution {
public:
    int dfs(int x, int y, const int &m,const int &n,vector<vector<int>>& matrix, vector<vector<int>>& dis) {
        if (dis[x][y]) return dis[x][y];
 
        for (int i = 0; i < 4; i++) {
            int
 
 nx = x + dx[i];
            int ny = y + dy[i];
            if (nx >= 0 && ny >= 0 && nx < m && ny < n && matrix[nx][ny] > matrix[x][y]) {
                dis[x][y] = max(dis[x][y], dfs(nx, ny, m, n, matrix, dis));
            }
        }
        return ++dis[x][y];
    }
 
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if (!matrix.size()) return 0;
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int> > dis(m, vector<int>(n, 0));
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                ans = max(ans, dfs( i, j, m, n, matrix, dis));
            }
        }
        return ans;
    }
};

Memoization-329. Longest Increasing Path in a Matrix