329. Longest Increasing Path in a Matrix

// Naive DFS Solution
// Time Limit Exceeded
public class Solution {
  private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
  private int m, n;

  public int longestIncreasingPath(int[][] matrix) {
      if (matrix.length == 0) return 0;
      m = matrix.length;
      n = matrix[0].length;
      int ans = 0;
      for (int i = 0; i < m; ++i)
          for (int j = 0; j < n; ++j)
              ans = Math.max(ans, dfs(matrix, i, j));
      return ans;
  }

  private int dfs(int[][] matrix, int i, int j) {
      int ans = 0;
      for (int[] d : dirs) {
          int x = i + d[0], y = j + d[1];
          if (0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j])
              ans = Math.max(ans, dfs(matrix, x, y));
      }
      return ++ans;
  }
}

Time Limit Exceeded

DFS를 생각했지만 탐리밋이 걸려버림 ㅠㅠ

// DFS + Memoization Solution
// Accepted and Recommended
public class Solution {
    private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    private int m, n;

    public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0) return 0;
        m = matrix.length; n = matrix[0].length;
        int[][] cache = new int[m][n];
        int ans = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                ans = Math.max(ans, dfs(matrix, i, j, cache));
        return ans;
    }

    private int dfs(int[][] matrix, int i, int j, int[][] cache) {
        if (cache[i][j] != 0) return cache[i][j];
        for (int[] d : dirs) {
            int x = i + d[0], y = j + d[1];
            if (0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j])
                cache[i][j] = Math.max(cache[i][j], dfs(matrix, x, y, cache));
        }
        return ++cache[i][j];
    }
}

Runtime: 8 ms, faster than 75.16% of Java online submissions for Longest Increasing Path in a Matrix.
Memory Usage: 39 MB, less than 93.84% of Java online submissions for Longest Increasing Path in a Matrix.

memoization이라고 한문장만 더해주면 런타임이 O(2^{m+n})에서 O(mn)으로 확 줌