Leetcode #1368: Minimum Cost to Make at Least One Valid Path in a Grid
In this guide, we solve Leetcode #1368 Minimum Cost to Make at Least One Valid Path in a Grid in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Graph, Array, Matrix, Shortest Path, Heap (Priority Queue)
Intuition
We need to repeatedly access the smallest or largest element as the input changes.
A heap provides fast insertions and removals while keeping order.
Approach
Push candidates into the heap as you scan, and pop when you need the best element.
Keep the heap size bounded if the problem requires a top-k structure.
Steps:
- Push candidates into a heap.
- Pop the best candidate when needed.
- Maintain heap size or invariants.
Example
Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3
Explanation: You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
The total cost = 3.
Python Solution
class Solution:
def minCost(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
dirs = [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]]
q = deque([(0, 0, 0)])
vis = set()
while q:
i, j, d = q.popleft()
if (i, j) in vis:
continue
vis.add((i, j))
if i == m - 1 and j == n - 1:
return d
for k in range(1, 5):
x, y = i + dirs[k][0], j + dirs[k][1]
if 0 <= x < m and 0 <= y < n:
if grid[i][j] == k:
q.appendleft((x, y, d))
else:
q.append((x, y, d + 1))
return -1
Complexity
The time complexity is O(n log n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.