Random Acyclic Maze Generator with given Entry and Exit point

Given two integers N and M, the task is to generate any N * M sized maze containing only 0 (representing a wall) and 1 (representing an empty space where one can move) with the entry point as P0 and exit point P1 and there is only one path between any two movable positions.

Note: P0 and P1 will be marked as 2 and 3 respectively and one can move through the moveable positions in 4 directions (up, down, right and left).

Examples:

Input: N = 5, M = 5, P0 = (0, 0), P1 = (4, 4)
Output: maze = [ [ 2 1 1 1 1 ], 
                             [ 1 0 1 0 1 ], 
                             [ 1 0 1 0 0 ], 
                             [ 1 1 0 1 0 ], 
                             [ 0 1 1 1 3 ] ]
Explanation: It is valid because there is no cycle,  
and there is no unreachable walkable position.
Some other options could be 
[ [ 2 1 1 1 1 ], 
  [ 1 0 1 0 1 ], 
  [ 1 0 1 0 0 ], 
  [ 1 1 1 1 0 ], 
  [ 0 0 0 1 3 ] ] 
or
[ [ 2 1 1 0 1 ], 
  [ 1 0 1 0 1 ], 
  [ 1 0 1 0 0 ], 
  [ 1 0 1 1 0 ], 
  [ 1 0 0 1 3 ] ].
But these are not valid because in the first one there is a cycle in the maze
and in the second one (0, 4) and (1, 4) cannot be reached from the starting point.

Approach: The problem can be solved based on the following idea:

Use a DFS which starts from the P0 position and moves to any of the neighbours but does not make a cycle and ends at P1. In this way, there will only be 1 path between any two movable positions.

Follow the below steps to implement the idea:

• Initialize a stack (S) for the iterative DFS, the matrix that will be returned as the random maze. 
• Insert the entry point P0 into the stack.
• While S is not empty, repeat the following steps:
  • Remove a position (say P) from S and mark it as seen.
    • If marking the position walkable, forms a cycle then don't include it as a moveable position.
    • Otherwise, set the position as walkable.
  • Insert the neighbours of P which are not visited in random order into the stack.
  • Random insertion in the stack guarantees that the maze being generated is random.
  • If any of the neighbours is the same as the P1 then insert it at the top so that we do not skip this position because of cycle formation.
• Mark the initial position P0 (with 2) and final position P1 (with 3)
• Return the maze.

Below is the implementation for the above approach:

[2, 0, 0, 1, 1]
[1, 1, 1, 0, 1]
[0, 0, 1, 1, 1]
[1, 1, 0, 0, 1]
[0, 1, 1, 1, 3]

Time Complexity: O(N * M)

• As the algorithm is basically a DFS with more conditions, the time complexity is the time complexity of the DFS: O(V+E) (where V is the number of vertices and E is the number of edges). 
• In this case, the number of vertices is the number of squares in the matrix: N * M. Each "vertex" (square) has at most 4 "edges" (4 adjacent squares) so E < 4*N*M. So O(V+E) will be O(5*N*M) i.e. O(N*M).

Auxiliary Space: O(N * M)

• Each auxiliary matrix needs O(N*M) of space. 
• The stack can't have more than N*M squares inside it, because it never holds (in this implementation) the same square more than one time. 
• The sum of all space mentioned will be O(N*M).