329. 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.

// DFS Recursion + Memoization
int helper(const vector<vector<int> >& matrix, int i, int j, vector<vector<int> >& memo, const vector<vector<int> >& dirs) {
    if (memo[i][j] != 0) return memo[i][j];
    int cur = 1;
    for (const vector<int>& dir : dirs) {
        int x = i + dir[0], y = j + dir[1];
        if (x < 0 || x >= matrix.size() || y < 0 || y >= matrix[0].size() || matrix[x][y] <= matrix[i][j]) continue;
        int len = helper(matrix, x, y, memo, dirs) + 1;
        cur = max(cur, len);
    }
    return memo[i][j] = cur;
}
int longestIncreasingPath(vector<vector<int>>& matrix) {
    if (matrix.empty()) return 0;
    int m = matrix.size(), n = matrix[0].size();
    vector<vector<int> > dirs({{1, 0}, {-1, 0}, {0, 1}, {0, -1}}), memo(m, vector<int>(n));
    int res = 1;
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            res = max(res, helper(matrix, i, j, memo, dirs));
        }
    }
    return res;
}