What is the time complexity of Minimum Cost to Make at Least One Valid Path in a Grid?

The optimal solution for Minimum Cost to Make at Least One Valid Path in a Grid runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the priority queue operations, while the space complexity is O(n) for storing the minimum costs.

What is the brute force solution for Minimum Cost to Make at Least One Valid Path in a Grid?

The brute force approach for Minimum Cost to Make at Least One Valid Path in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths from the starting cell to the destination. It checks each cell's sign and modifies it if necessary, calculating the total cost to reach the bottom-right cell. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Minimum Cost to Make at Least One Valid Path in a Grid?

Explain your thought process clearly while coding. Consider edge cases, such as grids with no modifications needed. Practice similar pathfinding problems to build intuition.

What are common mistakes when solving Minimum Cost to Make at Least One Valid Path in a Grid?

Not considering all possible directions when exploring neighbors. Failing to update costs correctly in the priority queue.

#1368

Minimum Cost to Make at Least One Valid Path in a Grid

Hard

ArrayBreadth-First SearchGraph TheoryHeap (Priority Queue)MatrixShortest PathGraph TraversalDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses a modified Dijkstra's algorithm to find the minimum cost path efficiently. It treats the grid as a graph where each cell connects to its neighbors, and the cost of moving to a neighbor depends on the sign in the current cell.

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Algorithm

3 steps

1Step 1: Initialize a priority queue to explore cells based on the cost to reach them, starting from (0, 0) with cost 0.
2Step 2: Use a 2D array to keep track of the minimum cost to reach each cell, initializing all values to infinity except the starting cell.
3Step 3: While the priority queue is not empty, extract the cell with the lowest cost, and explore its neighbors. Update the cost to reach each neighbor based on whether the sign points to it or not.

solution.py21 lines

1import heapq
2
3def minCost(grid):
4    m, n = len(grid), len(grid[0])
5    min_cost = [[float('inf')] * n for _ in range(m)]
6    min_cost[0][0] = 0
7    pq = [(0, 0, 0)]  # (cost, x, y)
8    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
9
10    while pq:
11        cost, x, y = heapq.heappop(pq)
12        if x == m - 1 and y == n - 1:
13            return cost
14        for i in range(4):
15            nx, ny = x + directions[i][0], y + directions[i][1]
16            if 0 <= nx < m and 0 <= ny < n:
17                new_cost = cost + (1 if grid[x][y] != i + 1 else 0)
18                if new_cost < min_cost[nx][ny]:
19                    min_cost[nx][ny] = new_cost
20                    heapq.heappush(pq, (new_cost, nx, ny))
21    return -1

ℹ

Complexity note: The time complexity is O(n log n) due to the priority queue operations, while the space complexity is O(n) for storing the minimum costs.

1Understanding the grid as a graph helps in visualizing the problem.
2Using a priority queue allows for efficient cost management.

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