How do you solve Minimum Cost Path with Alternating Directions II on LeetCode?

Minimum Cost Path with Alternating Directions II (LeetCode #3603) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming reduces redundant calculations.

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

The optimal solution for Minimum Cost Path with Alternating Directions II runs in O(m * n) time and uses O(m * n) space. The complexity comes from filling the DP table, iterating through each cell once.

What is the brute force solution for Minimum Cost Path with Alternating Directions II?

The brute force approach for Minimum Cost Path with Alternating Directions II is Brute Force, with O(n²) time complexity and O(1) space complexity. Try all possible paths from the start to the end cell, calculating the total cost for each path. This approach will explore every combination of moves and waits. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Minimum Cost Path with Alternating Directions II?

Minimum Cost Path with Alternating Directions II uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Grid Traversal.

What are the interview tips for Minimum Cost Path with Alternating Directions II?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar DP problems.

What are common mistakes when solving Minimum Cost Path with Alternating Directions II?

Forgetting to alternate between moving and waiting. Not initializing the DP table correctly.

#3603

Minimum Cost Path with Alternating Directions II

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingGrid Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

Use dynamic programming to store the minimum costs at each cell, alternating between moving and waiting. This avoids redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize a DP table to store minimum costs for each cell.
2Step 2: Fill the DP table by calculating costs based on previous cells and whether to wait or move.
3Step 3: Return the value in the bottom-right cell of the DP table.

solution.py12 lines

1def minCostPath(m, n, waitCost):
2    dp = [[float('inf')] * n for _ in range(m)]
3    dp[0][0] = 1
4    for i in range(m):
5        for j in range(n):
6            if i < m - 1:
7                dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + (i + 2) * (j + 1))
8            if j < n - 1:
9                dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + (i + 1) * (j + 2))
10            if i < m and j < n:
11                dp[i][j] += waitCost[i][j]
12    return dp[m - 1][n - 1]

ℹ

Complexity note: The complexity comes from filling the DP table, iterating through each cell once.

1Dynamic programming reduces redundant calculations.
2Understanding the cost structure is crucial for optimal pathfinding.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.