Leetcode #351: Android Unlock Patterns
In this guide, we solve Leetcode #351 Android Unlock Patterns 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
Android devices have a special lock screen with a 3 x 3 grid of dots. Users can set an "unlock pattern" by connecting the dots in a specific sequence, forming a series of joined line segments where each segment's endpoints are two consecutive dots in the sequence.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Bit Manipulation, Dynamic Programming, Backtracking, Bitmask
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: m = 1, n = 1
Output: 9
Python Solution
class Solution:
def numberOfPatterns(self, m: int, n: int) -> int:
def dfs(i: int, cnt: int = 1) -> int:
if cnt > n:
return 0
vis[i] = True
ans = int(cnt >= m)
for j in range(1, 10):
x = cross[i][j]
if not vis[j] and (x == 0 or vis[x]):
ans += dfs(j, cnt + 1)
vis[i] = False
return ans
cross = [[0] * 10 for _ in range(10)]
cross[1][3] = cross[3][1] = 2
cross[1][7] = cross[7][1] = 4
cross[1][9] = cross[9][1] = 5
cross[2][8] = cross[8][2] = 5
cross[3][7] = cross[7][3] = 5
cross[3][9] = cross[9][3] = 6
cross[4][6] = cross[6][4] = 5
cross[7][9] = cross[9][7] = 8
vis = [False] * 10
return dfs(1) * 4 + dfs(2) * 4 + dfs(5)
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.