Is Minimum Cost Path with Alternating Directions III easy or hard?

Minimum Cost Path with Alternating Directions III is categorized as Hard because the shortest path state includes more than just the current position. You need to combine graph traversal with additional direction-dependent constraints, which increases implementation complexity.

Minimum Cost Path with Alternating Directions III Python/Java solution

Python solutions usually rely on heapq for Dijkstra traversal, while Java implementations use PriorityQueue with custom state objects or arrays. Both versions maintain distances for each node-direction state and skip invalid transitions that break the alternating rule.

How to solve Minimum Cost Path with Alternating Directions III in O(n)?

An O(n)-style solution is possible only under restricted edge weights such as binary costs. In that case, 0-1 BFS replaces the priority queue with a deque and processes states in linear time relative to nodes and edges. General weighted graphs still require Dijkstra's algorithm.

What is the best approach for Minimum Cost Path with Alternating Directions III?

The best general approach is Dijkstra's algorithm on an expanded state graph. Each state stores both the current node and the previous movement direction so the alternating constraint can be enforced during traversal. This guarantees the minimum path cost for weighted graphs.

Is Minimum Cost Path with Alternating Directions III asked at Google/Amazon/Meta?

Alternating-direction shortest path problems commonly appear in interviews at large companies because they test graph modeling and state expansion techniques. Variants involving constrained BFS, Dijkstra, and direction-aware traversal are especially common at Google and Meta.

What data structure is used in Minimum Cost Path with Alternating Directions III?

The optimal solution typically uses a priority queue or min-heap together with an adjacency list representation of the graph. A distance table indexed by node and previous direction stores the best known cost for every traversal state.

What is the time complexity of Minimum Cost Path with Alternating Directions III?

The optimal Dijkstra-based solution runs in O((V * D + E) log(V * D)) time, where V is the number of nodes, E is the number of edges, and D is the number of direction states. Space complexity is O(V * D) because distance values are tracked per node-direction pair.

Description Hints 4 Solution Video 20:08

4003. Minimum Cost Path with Alternating Directions III

Hard61.4% acceptance

You are given two integers m and n representing the number of rows and columns of a grid. Your goal is to reach cell (m - 1, n - 1). You are also given a 2D integer array penalty.

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

You begin at cell (0, 0) and initially pay its entrance cost. Actions performed after entering (0, 0) are numbered starting from 1.

On each action, you may move to an adjacent cell or wait in the current cell. A move follows the parity rule if:

On an odd-numbered action, you move right or down.
On an even-numbered action, you move left or up.

The cost of an action is determined as follows:

If you move according to the parity rule, pay only the entrance cost of the destination cell.
If you move in a direction that violates the parity rule, pay the entrance cost of the destination cell plus penalty[i][j], where (i, j) is the cell you move from.
If you wait in cell (i, j), pay penalty[i][j].

After every move or wait, the action number increases by 1. Therefore, the required parity alternates after every action, regardless of whether a penalty was paid.

Return the minimum total cost required to reach (m - 1, n - 1).

Example 1:

Input: m = 2, n = 2, penalty = [[5,3],[1,4]]

Output: 8

Explanation:

The optimal path is:

Start at cell (0, 0) with entry cost (0 + 1) * (0 + 1) = 1.
Move 1: Move down to cell (1, 0) with entry cost (1 + 1) * (0 + 1) = 2.
Move 2: Move right to cell (1, 1) with entry cost (1 + 1) * (1 + 1) = 4 and an extra cost of penalty[1][0] = 1 for violating the even parity rule.

Thus, the total cost is 1 + 2 + 4 + 1 = 8.

Example 2:

Input: m = 2, n = 2, penalty = [[0,7],[3,2]]

Output: 7

Explanation:

The optimal path is:

Start at cell (0, 0) with entry cost (0 + 1) * (0 + 1) = 1.
Move 1: Wait at cell (0, 0) with an extra cost of penalty[0][0] = 0 to flip to even parity.
Move 2: Move right to cell (0, 1) with entry cost (0 + 1) * (1 + 1) = 2 and an extra cost of penalty[0][0] = 0 for violating the even parity rule.
Move 3: Move down to cell (1, 1) with entry cost (1 + 1) * (1 + 1) = 4.

Thus, the total cost is 1 + 0 + 2 + 0 + 4 = 7.

Example 3:

Input: m = 2, n = 3, penalty = [[8,0,9],[7,4,1]]

Output: 12

Explanation:

The optimal path is:

Start at cell (0, 0) with entry cost (0 + 1) * (0 + 1) = 1.
Move 1: Move right to cell (0, 1) with entry cost (0 + 1) * (1 + 1) = 2.
Move 2: Move right to cell (0, 2) with entry cost (0 + 1) * (2 + 1) = 3 and an extra cost of penalty[0][1] = 0 for violating the even parity rule.
Move 3: Move down to cell (1, 2) with entry cost (1 + 1) * (2 + 1) = 6.

Thus, the total cost is 1 + 2 + 3 + 0 + 6 = 12.

Constraints:

1 <= m, n <= 10⁵
2 <= m * n <= 10⁵
penalty.length == m
penalty[i].length == n
0 <= penalty[i][j] <= 10⁵

Topics

Array Graph Heap (Priority Queue)Matrix Shortest Path

Hints open one at a time

Hint 1

Treat each state as (i, j, parity), where parity indicates whether the next action is odd-numbered or even-numbered.

Hint 2

From each state, consider moving to every adjacent cell and also waiting. Every transition changes the parity.

Hint 3

The transition cost depends on whether the chosen move follows the current parity rule. Waiting costs penalty[i][j].

Hint 4

All transition costs are non-negative, so use Dijkstra's algorithm on these states. The answer is the minimum distance to the destination with either parity.

20:08

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1class Solution:
2    def minCost(self, m: int, n: int, penalty: List[List[int]]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

2
2
[[5,3],[1,4]]

Input

2
2
[[0,7],[3,2]]

Input

2
3
[[8,0,9],[7,4,1]]

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