Leetcode #752: Open the Lock
In this guide, we solve Leetcode #752 Open the Lock 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
You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Array, Hash Table, String
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202"
Output: 6
Explanation:
A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202".
Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid,
because the wheels of the lock become stuck after the display becomes the dead end "0102".
Python Solution
class Solution:
def openLock(self, deadends: List[str], target: str) -> int:
def next(s):
res = []
s = list(s)
for i in range(4):
c = s[i]
s[i] = '9' if c == '0' else str(int(c) - 1)
res.append(''.join(s))
s[i] = '0' if c == '9' else str(int(c) + 1)
res.append(''.join(s))
s[i] = c
return res
if target == '0000':
return 0
s = set(deadends)
if '0000' in s:
return -1
q = deque([('0000')])
s.add('0000')
ans = 0
while q:
ans += 1
for _ in range(len(q)):
p = q.popleft()
for t in next(p):
if t == target:
return ans
if t not in s:
q.append(t)
s.add(t)
return -1
Complexity
The time complexity is O(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.