Problem Description

You are given two integers m and n representing the number of rows and columns of a grid, respectively.

The cost to enter a cell (i, j) is defined as (i + 1) * (j + 1).

The path will always begin by entering cell (0, 0) on move 1 and paying the entrance cost.

At each step, you move to an adjacent cell, following an alternating pattern:

• On odd-numbered moves, you must move either right or down.
• On even-numbered moves, you must move either left or up.

Return the minimum total cost required to reach (m - 1, n - 1). If it is impossible, return -1.

For example, on a 2 x 2 grid, you start at (0, 0) and can only move right or down on your first move. Then, on your next move, you can only go left or up, and so on, always alternating directions as described above. Try to determine if it's possible to reach the bottom-right cell, and if so, calculate the smallest possible sum of the costs for entering each cell along that path.

Intuition

The unique movement rules mean you can only alternate between moving right/down and left/up on each step. After each move, you're forced to go in the opposite direction of what you just did, which makes it hard to make consistent progress toward the bottom-right corner.

If you try to reach (m - 1, n - 1) on any grid larger than either 1x1, 1x2, or 2x1, you'll find that the alternating pattern makes you get stuck or go back and forth without ever reaching the last cell. For small grids like 1x2 or 2x1, you can reach the end because there's only one possible path, and the alternating moves work out. But for larger grids, the restriction prevents you from ever reaching the target.

This is why only these specific small cases are possible, while all other grid sizes are impossible.

Solution Approach

This problem relies on observing that the movement constraints make most grid sizes impossible to reach (m-1, n-1) from (0, 0).

• If the grid is 1x1, you start at the only cell and are already at the target. The answer is the cost of (0,0), which is 1.
• For a 2x1 grid, the only path is: (0,0) (enter, pay 1), then move down to (1,0) (only move is down on the first step, pay 2). Total cost: 1 + 2 = 3.
• For a 1x2 grid, the only path is: (0,0) (enter, pay 1), then move right to (0,1) (only move is right on the first step, pay 2). Total cost: 1 + 2 = 3.
• Any larger grid cannot be completed due to the forced alternating movement pattern, which doesn't allow continuous progress toward the goal.

Because of this, the code checks for these three special cases and returns their sums, otherwise returns -1 for impossible situations:

if m == 1 and n == 1:
    return 1
if m == 2 and n == 1:
    return 3
if m == 1 and n == 2:
    return 3
return -1

No advanced algorithms or data structures are needed since the solution only requires checking these conditions.

Example Walkthrough

Let's walk through a 2 x 1 grid to illustrate the solution approach:

1. Starting Point: Begin at cell (0, 0).
   • Entrance cost: (0+1)*(0+1) = 1. Total cost so far: 1.
2. First Move: On the first move (odd-numbered), you can go either right or down according to the rules.
   • The only valid move in a 2x1 grid is down to (1, 0).
   • Entrance cost for (1, 0): (1+1)*(0+1) = 2. Total cost so far: 1 + 2 = 3.
3. Ending Point: You reach (1, 0), which is also the bottom-right cell.

There is no other possible path to reach the end within the movement constraints. The answer is 3.

If you attempt this on a 2 x 2 grid, you'll find that the alternating movement rules prevent reaching (1, 1), illustrating why only 1x1, 1x2, and 2x1 are valid cases.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(1) because it consists of a fixed number of conditional checks, regardless of the input values of m and n. The space complexity is also O(1) since no additional space proportional to the input size is used; all variables occupy only constant space.