Time and space complexity of recursion DFS when finding path in a maze

Question

The question is : a man is looking for a target(which is marked as 9) in a 2D array, where 0 represents walls and 1 represents roads. The method should find if the man can find the target or not.

I came up with the solution using DFS easily, but got stuck when trying to find out the time and space complexity of my code.

public boolean find(int[][] grid) {
    if(grid == null || grid.length == 0 || grid[0].length == 0 || grid[0][0] == 0) return 0;
    return helper(grid,0,0);    
}

private boolean helper(int[][] grid,int x, int y) {
    if(x >= 0 && x < grid.length && y >= 0 && y < grid[0].length) {
        if(grid[x][y] == 0) return false;
        else if(grid[x][y] == 9) return true;
        else if(grid[x][y] == 1) {
            grid[x][y]=2;
            return helper(grid,x,y-1) || helper(grid,x+1,y) || helper(grid,x,y+1) || helper(grid,x-1,y);
        }
        else return false;
    }
    else return false;
}

I think the time and space complexity is O(mn), but I am not sure.

Answer 1

Generally speaking, DFS has a time complexity of O(m + n) and a space complexity of O(n), where n is the number of locations you can be in and m is the total number of connections between locations (if you're familiar with graph theory, n is the number of nodes and m is the number of edges). In your case, if you have a grid whose size is w × h, then n = wh (there's one place you can be for each grid location) and m ≤ 4wh (since each location is adjacent to at most four other locations). This means that the runtime will be O(wh) and the space complexity will be O(wh) as well.