LeetCode 3596 - Minimum Cost Path with Alternating Directions I

This problem requires computing the minimum total cost to traverse a grid from the top-left corner (0, 0) to the bottom-right corner (m - 1, n - 1) while following a strict alternating movement pattern. Each cell (i, j) has a cost to enter, calculated as (i + 1) (j + 1).

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LeetCode Problem 3596

Difficulty: 🟡 Medium
Topics: Math, Brainteaser

Solution

Problem Understanding

This problem requires computing the minimum total cost to traverse a grid from the top-left corner (0, 0) to the bottom-right corner (m - 1, n - 1) while following a strict alternating movement pattern. Each cell (i, j) has a cost to enter, calculated as (i + 1) * (j + 1). Movement is constrained by the move number: on odd-numbered moves, you can go right or down, and on even-numbered moves, you can go left or up. The path begins at (0, 0) on move 1, paying its cost immediately.

The input consists of two integers, m and n, defining the number of rows and columns, with constraints up to $10^6$. This means a brute-force approach that explores all possible paths is infeasible due to combinatorial explosion. Edge cases include grids with a single row or column, as the alternating move pattern can make some destinations unreachable. The problem guarantees at least one cell in each dimension, so empty grids are not possible.

The key challenge is the alternating movement rule. For example, in a 2 x 2 grid, not every path that reaches the bottom-right corner is valid if it violates the alternation pattern.

Approaches

The brute-force approach would explore every possible path recursively, keeping track of the move number and summing the costs. At each cell, we consider all valid moves depending on whether the current move is odd or even, recursively computing the total cost to the destination. While this guarantees the correct answer, it is exponentially slow, roughly $O(2^{m+n})$, and infeasible for grids with m or n as large as $10^6$.

A key insight is that the movement alternation creates a repeating pattern, and the shortest path can be deduced without exploring every route. Specifically, due to the alternating pattern, the only reachable cells from (0, 0) follow a certain parity rule: the sum of the coordinates (i + j) determines whether a cell is reachable on odd or even moves. Using this insight, we can derive a mathematical formula for the minimum cost path without simulating every step. The optimal solution leverages the observation that to reach the bottom-right corner, the path must maximize the number of moves along the smaller dimension first, alternating directions as needed.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^(m+n)) O(m+n) Recursively explores all valid paths according to alternating move rules
Optimal O(1) O(1) Uses a mathematical observation based on grid size and move parity to directly compute minimum cost

Algorithm Walkthrough

  1. Check if the grid is 1 x 1. If yes, the minimum cost is simply (1 * 1) = 1.
  2. Check if either m or n is 1. In such cases, the path is a single line. Sum the costs along the line: for odd moves, move forward; for even moves, move backward. If the length exceeds 2, return -1 if the alternating rules make reaching the end impossible.
  3. For general grids with m > 1 and n > 1, observe that any cell can be reached by a combination of right/down moves followed by left/up moves. The minimum cost path in such cases will always include the starting cell, the bottom-right cell, and either the first row or first column cell adjacent to the start, depending on which dimension is larger.
  4. Using this observation, the minimum cost is computed as the sum of (0,0), (m-1,0), (0,n-1), and (m-1,n-1), adjusting for overlaps at the corners.
  5. Return the computed minimum cost.

Why it works: The invariant is that the alternating moves force the path to visit certain critical cells that minimize traversal cost. By leveraging the parity pattern of reachable cells, the algorithm guarantees that all valid paths are considered implicitly, without recursion. The problem requires finding the minimum cost to traverse a two-dimensional grid from the top-left corner (0, 0) to the bottom-right corner (m - 1, n - 1) under a non-standard movement constraint. Each cell (i, j) has a cost to enter defined as (i + 1) * (j + 1). Movement alternates based on the step number: odd-numbered moves allow movement right or down, while even-numbered moves allow movement left or up. The first move, entering (0, 0), is considered move 1.

The input consists of integers m and n, specifying the number of rows and columns. The output is the minimum total cost to reach the destination or -1 if it is impossible.

Constraints indicate that 1 <= m, n <= 10^6. This is crucial: naive solutions that explore all paths will be infeasible due to exponential growth, so a direct brute-force search is impossible. Notably, the problem allows m = 1 or n = 1, which could create situations where certain moves are impossible because the required alternating movement cannot be satisfied. For instance, a single-column grid cannot move right, which restricts odd moves, but downward moves may still be possible.

Key edge cases include:

  • m = 1 and n = 1, a 1x1 grid where the start is the destination.
  • Single-row or single-column grids, where horizontal or vertical movement is restricted.
  • Larger grids where the alternating movement forces "zig-zag" patterns, making certain cells unreachable.

Approaches

Brute Force

The brute-force approach would explore all possible paths using either recursion or backtracking. At each cell, we generate moves allowed by the parity of the current step and recurse, accumulating the cost. The algorithm returns the minimum total cost found. This approach is correct because it explicitly checks all valid paths, but it is prohibitively slow: the number of paths grows exponentially with the grid size (O(2^(m+n))), making it infeasible for m, n up to 10^6. Additionally, the recursion depth would exceed typical stack limits.

Optimal Approach

The key observation is that the alternating movement pattern creates a deterministic path structure. Odd moves allow only "forward" progress (right or down), while even moves allow "backward" movement (left or up). By simulating the parity of moves and using dynamic programming, we can compute the minimal cost in O(m * n) time with constant space optimization if needed.

Define two DP arrays:

  • dp_odd[i][j]: minimum cost to reach (i, j) on an odd-numbered move.
  • dp_even[i][j]: minimum cost to reach (i, j) on an even-numbered move.

Transition rules:

  • For dp_odd[i][j], the previous move must be even: dp_even[i-1][j] or dp_even[i][j-1] (if within bounds).
  • For dp_even[i][j], the previous move must be odd: dp_odd[i+1][j] or dp_odd[i][j+1] (if within bounds).

We iterate until all reachable cells are processed. The final answer is the minimum of dp_odd[m-1][n-1] and dp_even[m-1][n-1] (if defined).

Approach Time Complexity Space Complexity Notes
Brute Force O(2^(m+n)) O(m+n) Recursively explores all valid paths; infeasible for large grids
Optimal O(m * n) O(m * n) Uses DP with separate arrays for odd/even steps; scalable to given constraints

Algorithm Walkthrough

  1. Initialize dp_odd and dp_even arrays of size m x n with infinity. Set dp_odd[0][0] = 1 because the first move starts at (0, 0) with cost (0+1)*(0+1) = 1.
  2. Iterate over all cells in the grid in row-major order.
  3. For each cell (i, j), update dp_odd[i][j] by considering moves from dp_even[i-1][j] and dp_even[i][j-1] if i-1 >= 0 or j-1 >= 0.
  4. Similarly, update dp_even[i][j] by considering moves from dp_odd[i+1][j] and dp_odd[i][j+1] if i+1 < m or j+1 < n.
  5. Each update adds the entrance cost (i+1)*(j+1) to the previous cost.
  6. After filling both DP arrays, check if dp_odd[m-1][n-1] or dp_even[m-1][n-1] is finite. If both are infinite, return -1; otherwise, return the minimum.

Why it works: The algorithm maintains the invariant that dp_odd[i][j] and dp_even[i][j] always hold the minimum achievable cost to reach (i, j) on an odd or even move. All reachable paths respecting the movement alternation are considered exactly once in the transitions, guaranteeing correctness.

Python Solution

class Solution:
    def minCost(self, m: int, n: int) -> int:
        if m == 1 and n == 1:
            return 1
        if m == 1 or n == 1:
            if max(m, n) == 2:
                # Only two cells, directly reachable
                return m * n + 1
            return -1  # More than two cells in a line is unreachable due to alternation
        
        # General case for m > 1 and n > 1
        # Minimum cost includes start, end, and the two adjacent first row/column cells
        cost_start = 1  # (0,0)
        cost_end = m * n  # (m-1,n-1)
        cost_adjacent = (m) + (n)  # (m-1,0) + (0,n-1)
        return cost_start + cost_end + cost_adjacent

The Python solution handles the single-cell grid explicitly. For single-row or single-column grids, it checks if the path is feasible according to the alternating movement rules. For general grids, it calculates the minimum cost using the sum of the starting cell, ending cell, and minimal adjacent cells required by the move pattern. if m <= 0 or n <= 0: return -1

    inf = float('inf')
    dp_odd = [[inf] * n for _ in range(m)]
    dp_even = [[inf] * n for _ in range(m)]
    
    dp_odd[0][0] = 1  # cost of starting cell
    
    for i in range(m):
        for j in range(n):
            cost = (i + 1) * (j + 1)
            
            # Update odd move
            if i > 0:
                dp_odd[i][j] = min(dp_odd[i][j], dp_even[i-1][j] + cost)
            if j > 0:
                dp_odd[i][j] = min(dp_odd[i][j], dp_even[i][j-1] + cost)
            
            # Update even move
            if i + 1 < m:
                dp_even[i][j] = min(dp_even[i][j], dp_odd[i+1][j] + cost)
            if j + 1 < n:
                dp_even[i][j] = min(dp_even[i][j], dp_odd[i][j+1] + cost)
    
    result = min(dp_odd[m-1][n-1], dp_even[m-1][n-1])
    return result if result != inf else -1


**Explanation:** We initialize DP arrays for odd and even moves with infinity to indicate unreachable states. Each cell's cost is calculated as `(i+1)*(j+1)`. For each cell, we update the DP values according to possible previous moves allowed by parity. Finally, we return the minimum cost at the destination or `-1` if unreachable.

## Go Solution

```go
func minCost(m int, n int) int {
    if m == 1 && n == 1 {
        return 1
    }
    if m == 1 || n == 1 {
        if max(m, n) == 2 {
            return m*n + 1
        }
        return -1
    }
    costStart := 1
    costEnd := m * n
    costAdjacent := m + n
    return costStart + costEnd + costAdjacent
}

func max(a, b int) int {
    if a > b {
import "math"

func minCost(m int, n int) int {
    if m <= 0 || n <= 0 {
        return -1
    }
    
    inf := math.MaxInt64
    dpOdd := make([][]int, m)
    dpEven := make([][]int, m)
    for i := 0; i < m; i++ {
        dpOdd[i] = make([]int, n)
        dpEven[i] = make([]int, n)
        for j := 0; j < n; j++ {
            dpOdd[i][j] = inf
            dpEven[i][j] = inf
        }
    }
    
    dpOdd[0][0] = 1
    
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            cost := (i + 1) * (j + 1)
            
            if i > 0 && dpEven[i-1][j] != inf {
                dpOdd[i][j] = min(dpOdd[i][j], dpEven[i-1][j]+cost)
            }
            if j > 0 && dpEven[i][j-1] != inf {
                dpOdd[i][j] = min(dpOdd[i][j], dpEven[i][j-1]+cost)
            }
            if i+1 < m && dpOdd[i+1][j] != inf {
                dpEven[i][j] = min(dpEven[i][j], dpOdd[i+1][j]+cost)
            }
            if j+1 < n && dpOdd[i][j+1] != inf {
                dpEven[i][j] = min(dpEven[i][j], dpOdd[i][j+1]+cost)
            }
        }
    }
    
    result := min(dpOdd[m-1][n-1], dpEven[m-1][n-1])
    if result == inf {
        return -1
    }
    return result
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

In Go, the logic mirrors the Python version. A helper function max is added to check for the two-cell line case. Slice or array structures are unnecessary since the solution is purely formula-based.

Worked Examples

Example 1: m = 1, n = 1

  • Grid: [[1]]
  • Minimum cost: 1 (start equals end)

Example 2: m = 2, n = 1

  • Grid: [[1],[2]]
  • Move 1 (odd): start at (0,0), cost 1
  • Move 2 (even): move down to (1,0), cost 2
  • Total cost: 1 + 2 = 3

Example 3: m = 3, n = 3

  • Compute cost_start = 1, cost_end = 9, cost_adjacent = 3 + 3 = 6
  • Total cost: 1 + 9 + 6 = 16

Complexity Analysis

Measure Complexity Explanation
Time O(1) Direct computation using arithmetic formulas without loops
Space O(1) Only a few integer variables used, no additional data structures

This complexity is independent of the grid size due to the formula-based approach, which is necessary given the constraint m, n <= 10^6.

Test Cases

# Provided examples
assert Solution().minCost(1, 1) == 1  # single cell
assert Solution().minCost(2, 1) == 3  # single column, reachable

# Boundary cases
assert Solution().minCost(1, 2) == 3  # single row, reachable
assert Solution().minCost(1, 3) == -1  # single row, unreachable
assert Solution().minCost(2, 2) == 6  # minimal 2x2 grid
assert Solution().minCost(3, 3) == 16  # 3x3 grid general case
assert Solution().minCost(1000000, 1000000) == 1000000*1000000 + 1 + 1000000 + 1000000  # large grid
Test Why
1x1 Tests smallest grid, trivial path
2x1 Tests single-column, reachable path
1x3 Tests single-row, unreachable path due to alternation
2x2 Tests minimal square grid, general formula applies
3x3 Tests general square grid, larger than minimal
1_000_000 x 1_000_000 Tests large input, ensures O(1) computation

Edge Cases

First, a single-cell grid (1x1) must be handled explicitly, as it immediately satisfies the destination condition with minimal cost.

Second, single-row or single-column grids longer than two cells are often impossible to traverse due to alternating move rules. If the path is attempted naively, it may overshoot or move in the wrong direction. This is handled by returning -1 when max(m, n) > 2 and either dimension equals 1.

Third, large grids like m = n = 10^6 test the efficiency of the solution. Explanation: The Go implementation