[LeetCode] 1368. Minimum Cost to Make at Least One Valid Path in a Grid (Python)

 

 

난이도: Hard

문제 설명

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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. The sign of grid[i][j] can be:

• 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
• 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
• 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
• 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some signs on the cells of the grid that point outside the grid.

 

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

 

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

 

Return the minimum cost to make the grid have at least one valid path.

 

문제 예제

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Example 1:

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.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0
Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

Input: grid = [[1,2],[4,3]]
Output: 1

 

제한 사항

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• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• 1 <= grid[i][j] <= 4

 

✏️ Solution(솔루션)

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])

        visited = defaultdict(bool)
        visited[(0,0,0)] = True
        moves = [[], [0, 1], [0, -1], [1,0], [-1,0]]
        q = [(0,0,0)] 
        heapq.heapify(q)

        while q:
            cost, x, y = heapq.heappop(q)
            
            if x == n-1 and y == m-1:
                return cost

            for i in range(1, 5):
                nx = x + moves[i][0]
                ny = y + moves[i][1]
                if 0<= nx < n and 0<=ny<m:
                    if i == grid[x][y]:
                        if not visited[(cost, nx, ny)]:
                            visited[(cost, nx, ny)] = True
                            heapq.heappush(q, (cost, nx, ny))
                    else:
                        if not visited[(cost+1, nx, ny)]:
                            visited[(cost+1, nx, ny)] = True
                            heapq.heappush(q, (cost+1, nx, ny))

 

 

나는 이 문제를 bfs로 해결했다.

 

0, 0 좌표에서 시작해서 상하좌우로 모두 움직여 만약 grid[x][y]와 미리 만들어둔 moves의 인덱스와 같다면 cost를 그대로 두고 아니라고 한다면 cost에 +1을 해주어 관리했다.

 

그리고 문제는 최소한의 cost를 return해야하기 때문에 deque를 사용하는 것이 아닌 heapq를 사용해서 작은 cost를 갖는 애들로 먼저 연산을 수행했다.

 

연산을 계속 수행하다가 q에서 pop한 x,y값이 n-1, m-1이면 cost를 반환하고 정답을 맞출 수 있었다.

 

이 코드는 수행시간이 다른사람들에 비해 오래걸리긴 했다. 좀 더 효율적이게 한번 더 짜봐야할 것 같다.

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문제: 1368. Minimum Cost to Make at Least One Valid Path in a Grid

 

깃허브: github

algorithmPractice/LeetCode/1485-minimum-cost-to-make-at-least-one-valid-path-in-a-grid at main · laewonJeong/algorithmPractice