How do you solve Minimum Cost Path with Teleportations on LeetCode?

Minimum Cost Path with Teleportations (LeetCode #3651) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n * k) time complexity and O(m * n * k) space complexity. Key insight: Dynamic programming optimizes the exploration of paths.

What is the time complexity of Minimum Cost Path with Teleportations?

The optimal solution for Minimum Cost Path with Teleportations runs in O(m * n * k) time and uses O(m * n * k) space. The complexity arises from filling a 3D DP table, where each cell is evaluated for each teleportation count.

What is the brute force solution for Minimum Cost Path with Teleportations?

The brute force approach for Minimum Cost Path with Teleportations is Brute Force, with O(n²) time complexity and O(1) space complexity. Explore all possible paths from the top-left to the bottom-right cell, considering both normal moves and teleportations. This method evaluates every potential route, leading to high computational costs. The optimal Optimal Solution approach improves this to O(m * n * k).

What algorithm or data structure is used to solve Minimum Cost Path with Teleportations?

Minimum Cost Path with Teleportations uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: .

What are common mistakes when solving Minimum Cost Path with Teleportations?

Forgetting to account for teleportation limits.

#3651

Minimum Cost Path with Teleportations

Hard

ArrayDynamic ProgrammingMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

Utilize dynamic programming to systematically calculate the minimum cost to reach each cell, incorporating both normal moves and teleportations. This approach avoids redundant calculations and efficiently tracks costs.

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Algorithm

3 steps

1Step 1: Initialize a DP table with dimensions (m x n x (k + 1)) to store minimum costs.
2Step 2: Fill the DP table by considering normal moves and teleportations from each cell.
3Step 3: Return the minimum cost found at the bottom-right cell.

solution.py22 lines

1def minCost(grid, k):
2    from collections import deque
3    m, n = len(grid), len(grid[0])
4    dp = [[[float('inf')] * (k + 1) for _ in range(n)] for _ in range(m)]
5    dp[0][0][0] = 0
6    q = deque([(0, 0, 0)])
7    while q:
8        x, y, t = q.popleft()
9        for dx, dy in [(1, 0), (0, 1)]:
10            nx, ny = x + dx, y + dy
11            if 0 <= nx < m and 0 <= ny < n:
12                cost = dp[x][y][t] + grid[nx][ny]
13                if cost < dp[nx][ny][t]:
14                    dp[nx][ny][t] = cost
15                    q.append((nx, ny, t))
16        if t < k:
17            for i in range(m):
18                for j in range(n):
19                    if grid[i][j] <= grid[x][y] and dp[x][y][t] < dp[i][j][t + 1]:
20                        dp[i][j][t + 1] = dp[x][y][t]
21                        q.append((i, j, t + 1))
22    return min(dp[m - 1][n - 1])

ℹ

Complexity note: The complexity arises from filling a 3D DP table, where each cell is evaluated for each teleportation count.

1Dynamic programming optimizes the exploration of paths.
2Teleportation allows skipping costly cells, reducing overall cost.

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