Leetcode #773: Sliding Puzzle
In this guide, we solve Leetcode #773 Sliding Puzzle 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
On an 2 x 3 board, there are five tiles labeled from 1 to 5, and an empty square represented by 0. A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Breadth-First Search, Memoization, Array, Dynamic Programming, Backtracking, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.
Python Solution
class Solution:
def slidingPuzzle(self, board: List[List[int]]) -> int:
t = [None] * 6
def gets():
for i in range(2):
for j in range(3):
t[i * 3 + j] = str(board[i][j])
return ''.join(t)
def setb(s):
for i in range(2):
for j in range(3):
board[i][j] = int(s[i * 3 + j])
def f():
res = []
i, j = next((i, j) for i in range(2) for j in range(3) if board[i][j] == 0)
for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
x, y = i + a, j + b
if 0 <= x < 2 and 0 <= y < 3:
board[i][j], board[x][y] = board[x][y], board[i][j]
res.append(gets())
board[i][j], board[x][y] = board[x][y], board[i][j]
return res
start = gets()
end = "123450"
if start == end:
return 0
vis = {start}
q = deque([(start)])
ans = 0
while q:
ans += 1
for _ in range(len(q)):
x = q.popleft()
setb(x)
for y in f():
if y == end:
return ans
if y not in vis:
vis.add(y)
q.append(y)
return -1
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.