Problem Description

You have a grid with m rows and n columns. Each cell (i, j) in the grid has an entrance cost of (i + 1) * (j + 1).

You start at cell (0, 0) on move 1, paying its entrance cost of 1 * 1 = 1.

From any position, you can move to adjacent cells (up, down, left, or right), but your movement direction depends on the move number:

• Odd-numbered moves (1st, 3rd, 5th, etc.): You can only move right or down
• Even-numbered moves (2nd, 4th, 6th, etc.): You can only move left or up

Your goal is to reach the bottom-right cell (m - 1, n - 1) with the minimum total cost. The total cost is the sum of entrance costs for all cells you visit along your path.

Return the minimum total cost to reach the destination. If it's impossible to reach (m - 1, n - 1), return -1.

The key insight is that the alternating movement pattern severely restricts possible paths. Starting from (0, 0), you move right/down on odd moves and left/up on even moves. This creates a back-and-forth pattern that makes it impossible to reach most destinations. Only very specific grid sizes allow reaching the bottom-right corner:

• A 1 × 1 grid (already at destination)
• A 2 × 1 grid (move down on move 1)
• A 1 × 2 grid (move right on move 1)

All other grid sizes result in impossible paths due to the alternating movement constraints.

Intuition

Let's trace through what happens when we try to navigate this grid with the alternating movement pattern.

Starting at (0, 0), on move 1 (odd), we can only go right to (0, 1) or down to (1, 0). Then on move 2 (even), we must go left or up, which brings us back toward or exactly to where we started. On move 3 (odd), we go right or down again, and on move 4 (even), we go left or up again.

This creates a fundamental problem: we're essentially taking one step forward and one step back repeatedly. The alternating pattern means we can never make consistent progress toward the bottom-right corner.

Let's analyze when we can actually reach (m-1, n-1):

1. If the grid is 1 × 1: We start at (0, 0) which is already the destination. Cost = 1.

2. If the grid is 2 × 1: From (0, 0), we move down to (1, 0) on move 1 (odd move allows down). We've reached the destination! Cost = 1 + 2 = 3.

3. If the grid is 1 × 2: From (0, 0), we move right to (0, 1) on move 1 (odd move allows right). We've reached the destination! Cost = 1 + 2 = 3.

4. For any larger grid: Let's see why it's impossible. If we need to go both right AND down to reach (m-1, n-1), we face a contradiction. After moving right or down on an odd move, the next even move forces us back (left or up). We can never accumulate progress in both directions simultaneously. The furthest we can ever get from the origin is one step in a single direction before being pulled back.

The movement pattern essentially confines us to a very small region around the starting point, making it impossible to reach any cell that requires moving more than one step away from (0, 0) in total Manhattan distance.

Learn more about Math patterns.

Solution Approach

The solution is remarkably straightforward once we understand that only three specific grid configurations allow us to reach the destination.

The implementation simply checks for these three valid cases:

1. Check if it's a 1 × 1 grid (m == 1 and n == 1):

   • We're already at the destination (0, 0)
   • Return cost = 1

2. Check if it's a 2 × 1 grid (m == 2 and n == 1):

   • We can reach (1, 0) in one move by going down
   • Total cost = 1 (starting cell) + 2 (destination cell) = 3
   • Return cost = 3

3. Check if it's a 1 × 2 grid (m == 1 and n == 2):

   • We can reach (0, 1) in one move by going right
   • Total cost = 1 (starting cell) + 2 (destination cell) = 3
   • Return cost = 3

4. For all other grid sizes:

   • It's impossible to reach the destination due to the alternating movement constraints
   • Return -1

The solution doesn't require any complex algorithms, data structures, or dynamic programming. It's a direct mathematical observation that the alternating movement pattern makes it impossible to reach any cell beyond these three specific cases.

The time complexity is O(1) since we're only performing constant-time comparisons. The space complexity is also O(1) as we don't use any additional data structures.

Example Walkthrough

Let's walk through a few examples to understand why only specific grid sizes work:

Example 1: 3×3 Grid (Impossible)

Starting at (0,0) with cost 1:

• Move 1 (odd): Can go right to (0,1) or down to (1,0)
  • Let's go right to (0,1), pay cost 2. Total: 1+2=3
• Move 2 (even): Must go left or up
  • Can only go left back to (0,0), pay cost 1. Total: 3+1=4
• Move 3 (odd): Can go right or down again
  • Go down to (1,0), pay cost 2. Total: 4+2=6
• Move 4 (even): Must go left or up
  • Go up back to (0,0), pay cost 1. Total: 6+1=7

We're stuck in a loop! We can never reach (2,2) because:

• To reach (2,2), we need to move 2 steps right AND 2 steps down
• But after each forward move (odd), we're forced backward (even)
• We can never accumulate progress in both directions

Example 2: 2×1 Grid (Possible)

Starting at (0,0) with cost 1:

• Move 1 (odd): Can go down to (1,0)
  • Move down to (1,0), pay cost 2. Total: 1+2=3
• We've reached the destination (1,0)!

Example 3: 1×2 Grid (Possible)

Starting at (0,0) with cost 1:

• Move 1 (odd): Can go right to (0,1)
  • Move right to (0,1), pay cost 2. Total: 1+2=3
• We've reached the destination (0,1)!

Example 4: 2×2 Grid (Impossible)

Starting at (0,0) with cost 1, we need to reach (1,1):

• Move 1 (odd): Go right to (0,1), pay cost 2. Total: 3
• Move 2 (even): Must go left or up
  • Can only go left back to (0,0), pay cost 1. Total: 4
• Move 3 (odd): Go down to (1,0), pay cost 2. Total: 6
• Move 4 (even): Must go up back to (0,0), pay cost 1. Total: 7

We can never reach (1,1) which requires being 1 step right AND 1 step down simultaneously. The alternating pattern prevents us from maintaining progress in multiple directions.

The pattern is clear: only grids where the destination is reachable in exactly one move (2×1 or 1×2) or where we're already at the destination (1×1) are solvable.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function only performs a fixed number of constant-time operations regardless of the input values. It checks a few conditional statements and returns a value directly without any loops or recursive calls. The number of operations remains the same whether m and n are small or large values.

The space complexity is O(1) because the function uses only a constant amount of extra space. No additional data structures are created, and the space usage doesn't scale with the input size. The function only uses the input parameters m and n and returns a single integer value.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Movement Pattern

Pitfall: Developers often misinterpret the alternating movement rules, thinking they can eventually reach any cell by zigzagging through the grid. They might attempt to implement complex pathfinding algorithms like BFS or dynamic programming.

Why it happens: The problem statement mentions finding the "minimum cost," which typically suggests an optimization problem requiring algorithms like Dijkstra's or DP. This leads to overthinking the solution.

Solution: Carefully trace through the movement pattern manually:

• Move 1 (odd): From (0,0), can only go right or down
• Move 2 (even): Can only go left or up (back towards origin)
• Move 3 (odd): Can only go right or down again

This reveals that you're essentially oscillating and can never make consistent progress toward distant cells.

2. Incorrect Cost Calculation

Pitfall: When calculating the cost for valid cases, developers might:

• Forget to include the starting cell's cost
• Use 0-based indexing for cost calculation instead of the specified (i+1) * (j+1) formula
• Calculate cost[1][0] = 1 * 0 = 0 instead of (1+1) * (0+1) = 2

Solution: Remember that:

• Cell (i,j) has cost = (i+1) * (j+1)
• Starting cell (0,0) has cost = 1 * 1 = 1
• Cell (1,0) has cost = 2 * 1 = 2
• Cell (0,1) has cost = 1 * 2 = 2

3. Attempting to Generalize the Pattern

Pitfall: Some might try to find a formula or pattern for larger grids, thinking there could be valid paths for grids like 3×3 or 2×2.

Why it happens: The restriction to only three valid cases seems too limiting, leading developers to search for additional valid configurations.

Solution: Mathematically prove why only these three cases work:

• From (0,0), after an odd move you're at most 1 step away
• After an even move, you must move back toward (0,0)
• This creates a maximum reachable distance of 1 in any direction
• Only cells within this 1-step radius from origin can be destinations

4. Over-engineering the Solution

Pitfall: Implementing unnecessary data structures or algorithms:

# Overcomplicated approach - DON'T DO THIS
def minCost(self, m: int, n: int) -> int:
    visited = set()
    queue = deque([(0, 0, 1, 1)])  # row, col, move_num, cost
    # ... complex BFS implementation ...

Solution: Recognize that this is a constraint analysis problem, not a pathfinding problem. The solution is deterministic based on grid dimensions alone - no search needed.

5. Missing Edge Cases in Testing

Pitfall: Not testing all three valid cases thoroughly, or assuming symmetric behavior (e.g., if 2×1 works, then 1×2 must work the same way).

Solution: Explicitly test each valid configuration:

• Test m=1, n=1 → returns 1
• Test m=2, n=1 → returns 3
• Test m=1, n=2 → returns 3
• Test m=2, n=2 → returns -1 (impossible case)