{ "data": { "question": { "questionId": "3870", "questionFrontendId": "3568", "categoryTitle": "Algorithms", "boundTopicId": 3688624, "title": "Minimum Moves to Clean the Classroom", "titleSlug": "minimum-moves-to-clean-the-classroom", "content": "
You are given an m x n
grid classroom
where a student volunteer is tasked with cleaning up litter scattered around the room. Each cell in the grid is one of the following:
'S'
: Starting position of the student'L'
: Litter that must be collected (once collected, the cell becomes empty)'R'
: Reset area that restores the student's energy to full capacity, regardless of their current energy level (can be used multiple times)'X'
: Obstacle the student cannot pass through'.'
: Empty spaceYou are also given an integer energy
, representing the student's maximum energy capacity. The student starts with this energy from the starting position 'S'
.
Each move to an adjacent cell (up, down, left, or right) costs 1 unit of energy. If the energy reaches 0, the student can only continue if they are on a reset area 'R'
, which resets the energy to its maximum capacity energy
.
Return the minimum number of moves required to collect all litter items, or -1
if it's impossible.
\n
Example 1:
\n\nInput: classroom = ["S.", "XL"], energy = 2
\n\nOutput: 2
\n\nExplanation:
\n\n(0, 0)
with 2 units of energy.(1, 0)
contains an obstacle 'X', the student cannot move directly downward.(0, 0)
→ (0, 1)
with 1 unit of energy and 1 unit remaining.(0, 1)
→ (1, 1)
to collect the litter 'L'
.Example 2:
\n\nInput: classroom = ["LS", "RL"], energy = 4
\n\nOutput: 3
\n\nExplanation:
\n\n(0, 1)
with 4 units of energy.(0, 1)
→ (0, 0)
to collect the first litter 'L'
with 1 unit of energy used and 3 units remaining.(0, 0)
→ (1, 0)
to 'R'
to reset and restore energy back to 4.(1, 0)
→ (1, 1)
to collect the second litter 'L'
.Example 3:
\n\nInput: classroom = ["L.S", "RXL"], energy = 3
\n\nOutput: -1
\n\nExplanation:
\n\nNo valid path collects all 'L'
.
\n
Constraints:
\n\n1 <= m == classroom.length <= 20
1 <= n == classroom[i].length <= 20
classroom[i][j]
is one of 'S'
, 'L'
, 'R'
, 'X'
, or '.'
1 <= energy <= 50
'S'
in the grid.'L'
cells in the grid.给你一个 m x n
的网格图 classroom
,其中一个学生志愿者负责清理散布在教室里的垃圾。网格图中的每个单元格是以下字符之一:
'S'
:学生的起始位置'L'
:必须收集的垃圾(收集后,该单元格变为空白)'R'
:重置区域,可以将学生的能量恢复到最大值,无论学生当前的能量是多少(可以多次使用)'X'
:学生无法通过的障碍物'.'
:空白空间同时给你一个整数 energy
,表示学生的最大能量容量。学生从起始位置 'S'
开始,带着 energy
的能量出发。
每次移动到相邻的单元格(上、下、左或右)会消耗 1 单位能量。如果能量为 0,学生此时只有处在 'R'
格子时可以继续移动,此区域会将能量恢复到 最大 能量值 energy
。
返回收集所有垃圾所需的 最少 移动次数,如果无法完成,返回 -1
。
\n\n
示例 1:
\n\n输入: classroom = [\"S.\", \"XL\"], energy = 2
\n\n输出: 2
\n\n解释:
\n\n(0, 0)
开始,带着 2 单位的能量。(1, 0)
有一个障碍物 'X',学生无法直接向下移动。(0, 0)
→ (0, 1)
,消耗 1 单位能量,剩余 1 单位。(0, 1)
→ (1, 1)
,收集垃圾 'L'
。示例 2:
\n\n输入: classroom = [\"LS\", \"RL\"], energy = 4
\n\n输出: 3
\n\n解释:
\n\n(0, 1)
开始,带着 4 单位的能量。(0, 1)
→ (0, 0)
,收集第一个垃圾 'L'
,消耗 1 单位能量,剩余 3 单位。(0, 0)
→ (1, 0)
,到达 'R'
重置区域,恢复能量为 4。(1, 0)
→ (1, 1)
,收集第二个垃圾 'L'
。示例 3:
\n\n输入: classroom = [\"L.S\", \"RXL\"], energy = 3
\n\n输出: -1
\n\n解释:
\n\n没有有效路径可以收集所有 'L'
。
\n\n
提示:
\n\n1 <= m == classroom.length <= 20
1 <= n == classroom[i].length <= 20
classroom[i][j]
是 'S'
、'L'
、'R'
、'X'
或 '.'
之一1 <= energy <= 50
'S'
。'L'
单元格。(x, y, mask, e, steps)
, initializing with (sx, sy, 0, energy, 0)
, and for each move update e
(–1 per step), update mask
on 'L', reset e=energy
on 'R', and return steps
when mask == fullMask
.",
"Maintain a 3D array bestEnergy[x][y][mask]
storing the maximum e
seen for each (x,y,mask)
and skip any new state with e <= bestEnergy[x][y][mask]
to prune."
],
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