Problem Description

This problem is about counting valid unlock patterns on an Android lock screen. The lock screen has a 3×3 grid of dots numbered 1-9 (arranged like a phone keypad):

1 2 3
4 5 6
7 8 9

An unlock pattern is a sequence of connected dots where:

1. All dots must be distinct - each dot can only be used once in a pattern
2. No jumping through unvisited dots - if a line between two consecutive dots passes through the center of another dot, that middle dot must have been visited earlier in the sequence

For example:

• Connecting dots 1→3 requires dot 2 to be visited first (since the line passes through dot 2's center)
• Connecting dots 1→9 requires dot 5 to be visited first (diagonal through center)
• Connecting dots 2→9 is allowed without visiting dot 5 first (the line doesn't pass through dot 5's center)

The key pairs that require a middle dot to be visited first are:

• Horizontal: 1↔3 (through 2), 4↔6 (through 5), 7↔9 (through 8)
• Vertical: 1↔7 (through 4), 2↔8 (through 5), 3↔9 (through 6)
• Diagonal: 1↔9 (through 5), 3↔7 (through 5)

Given two integers m and n, you need to find the total number of unique valid unlock patterns that have at least m dots and at most n dots.

The solution uses DFS (depth-first search) to explore all possible valid patterns. It maintains a cross table to track which middle dot must be visited when connecting two dots, and uses symmetry to optimize the calculation (patterns starting from corners are 4-fold symmetric, patterns starting from edges are 4-fold symmetric, and the center is unique).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: While the 3×3 grid could be viewed as a graph, the problem is fundamentally about generating valid sequences/patterns, not about graph traversal or graph properties.

Need to solve for kth smallest/largest?

• No: We're counting all valid patterns, not finding a specific kth element.

Involves Linked Lists?

• No: The problem doesn't involve linked list data structures.

Does the problem have small constraints?

• Yes: The grid is fixed at 3×3 (only 9 dots), and patterns have length between m and n where both are at most 9. This is a very small search space.

Brute force / Backtracking?

• Yes: Given the small constraints and the need to explore all possible valid sequences while respecting complex validity rules (no jumping through unvisited dots), backtracking is the natural approach. We need to:
  • Try each possible next dot
  • Check if it's valid (not visited and no illegal jumps)
  • Recursively explore from that dot
  • Backtrack by unmarking the dot when done

Conclusion: The flowchart correctly leads us to use a Backtracking approach. The solution implements this by:

1. Using DFS to explore all possible patterns starting from each dot
2. Maintaining a vis array to track visited dots (and backtracking by unmarking them)
3. Using a cross table to check if moves require an intermediate dot to be visited first
4. Leveraging symmetry (corner dots, edge dots, center dot) to optimize the counting

Intuition

When we think about this problem, we're essentially trying to count all possible ways to draw patterns on the lock screen. Since we need to explore all valid sequences and each choice depends on what we've chosen before (visited dots affect future valid moves), this naturally points to a backtracking approach.

The key insight is recognizing that certain moves are restricted. When connecting two dots, if the straight line passes through another dot's center, that dot must be visited first. For instance, going from dot 1 to dot 3 requires dot 2 to be visited because the line passes directly through it. This creates dependencies between moves.

To handle these dependencies efficiently, we can precompute which dot (if any) must be visited when moving between any two dots. This leads to creating a cross table where cross[i][j] tells us which dot must be visited before we can move from dot i to dot j. Most moves don't have restrictions (like 1→2 or 2→9), so most entries in this table are 0.

The next crucial observation is about symmetry. The 3×3 grid has rotational and reflective symmetry:

• Corner dots (1, 3, 7, 9) are all equivalent - patterns starting from any corner can be rotated to match patterns from other corners
• Edge dots (2, 4, 6, 8) are similarly equivalent
• The center dot (5) is unique

This means instead of starting our search from all 9 positions, we only need to compute patterns starting from:

• One corner (multiply result by 4)
• One edge (multiply result by 4)
• The center (use result as is)

The backtracking algorithm then becomes straightforward: from each starting position, we try to extend the pattern to every unvisited dot that we can legally reach (either directly or when the required intermediate dot has been visited). We count valid patterns when their length falls within the [m, n] range, and backtrack by unmarking dots after exploring all possibilities from them.

Solution Approach

The solution implements a depth-first search with backtracking to explore all valid unlock patterns. Let's walk through the implementation step by step:

1. Setting up the Cross Table

First, we create a 10×10 table (using indices 1-9, ignoring index 0) to store which dot must be visited when moving between two dots:

cross = [[0] * 10 for _ in range(10)]
cross[1][3] = cross[3][1] = 2  # Moving between 1 and 3 requires 2
cross[1][7] = cross[7][1] = 4  # Moving between 1 and 7 requires 4
cross[1][9] = cross[9][1] = 5  # Moving between 1 and 9 requires 5
# ... and so on for all restricted moves

The table is symmetric since the restriction works both ways. Most entries remain 0, meaning no intermediate dot is required.

2. DFS with Backtracking

The core algorithm is implemented in the dfs function:

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

The function works as follows:

• i is the current dot position, cnt is the current pattern length
• If pattern length exceeds n, return 0 (invalid pattern)
• Mark current dot as visited: vis[i] = True
• If current length is within [m, n], count this as a valid pattern: ans = int(cnt >= m)
• Try extending to each unvisited dot j:
  • Check if move is valid: either no intermediate dot required (x == 0) or the intermediate dot is already visited (vis[x])
  • Recursively explore from dot j with incremented count
• Backtrack by unmarking the current dot: vis[i] = False
• Return total count of valid patterns found

3. Exploiting Symmetry

Instead of starting DFS from all 9 positions, we leverage symmetry:

return dfs(1) * 4 + dfs(2) * 4 + dfs(5)

• dfs(1) * 4: Patterns starting from corner (1, 3, 7, 9 are equivalent)
• dfs(2) * 4: Patterns starting from edge (2, 4, 6, 8 are equivalent)
• dfs(5): Patterns starting from center (unique position)

4. Visited Array

The vis array tracks which dots have been used in the current pattern:

vis = [False] * 10

This array is crucial for:

• Preventing revisiting dots (ensuring all dots are distinct)
• Checking if required intermediate dots have been visited
• Backtracking by resetting the visited status

The time complexity is O(9!) in the worst case (exploring all permutations), but practically much less due to early termination and constraints. The space complexity is O(1) for the fixed-size data structures plus O(n) for the recursion stack depth.

Example Walkthrough

Let's walk through finding all valid patterns of length 2-3 (m=2, n=3) to illustrate the solution approach.

Setup:

• Create a cross table where cross[i][j] stores the dot that must be visited before moving from i to j
• Initialize vis array to track visited dots

Starting from Corner (dot 1):

1. Mark dot 1 as visited: vis[1] = True

2. Since length is 1 (< m=2), don't count this as valid yet

3. Try extending to each unvisited dot:

   • 1→2: No restriction (cross[1][2] = 0), valid move
     • Pattern [1,2]: length=2, count as valid (✓)
     • Try extending: 2→3, 2→4, 2→5, 2→6, 2→7, 2→8, 2→9 all valid
     • Each creates a length-3 pattern like [1,2,3], [1,2,4], etc. (✓)
   • 1→3: Requires dot 2 (cross[1][3] = 2), but dot 2 not visited, skip
   • 1→4: No restriction, valid move
     • Pattern [1,4]: length=2 (✓)
     • Extend to get patterns like [1,4,2], [1,4,5], [1,4,7], etc.
   • 1→5: No restriction, valid move
     • Pattern [1,5]: length=2 (✓)
     • Can now reach 3,7,9 through center: [1,5,3], [1,5,7], [1,5,9]
   • 1→7: Requires dot 4 (cross[1][7] = 4), dot 4 not visited, skip
   • 1→9: Requires dot 5 (cross[1][9] = 5), dot 5 not visited, skip

4. Backtrack: vis[1] = False

Key observations from this partial exploration:

• Direct neighbors (like 1→2, 1→4) are always valid
• Restricted moves (1→3, 1→7, 1→9) become available only after visiting the middle dot
• Once we visit dot 5, diagonal moves like 1→9 become possible in subsequent steps

Starting from Edge (dot 2): Similar process, but dot 2 has different connectivity:

• Can reach 1,3,4,5,6,7,8,9 directly (no restrictions from dot 2)
• More flexibility in creating patterns

Starting from Center (dot 5):

• Can reach all 8 surrounding dots directly
• Maximum flexibility, no restricted moves from center

Final Calculation:

Total = dfs(1) × 4 + dfs(2) × 4 + dfs(5)

For our example with m=2, n=3:

• Patterns from corner (×4 for symmetry)
• Patterns from edge (×4 for symmetry)
• Patterns from center (×1, unique)

This symmetry optimization reduces computation from 9 starting points to just 3, making the algorithm much more efficient while still counting all valid patterns correctly.

Solution Implementation

Time and Space Complexity

Time Complexity: O(9!)

The algorithm uses DFS to explore all possible pattern combinations. In the worst case (when n = 9), we need to explore all possible permutations of placing digits 1-9 in sequence.

• From each position, we can potentially visit up to 9 different next positions
• The maximum depth of recursion is n
• At each level, we check all 9 positions, leading to a branching factor that decreases as more positions are visited
• The upper bound is 9 * 8 * 7 * ... * 1 = 9! possible paths to explore
• The algorithm is called 9 times total (4 times for corner positions, 4 times for edge positions, 1 time for center), but this doesn't change the factorial nature of the complexity

Space Complexity: O(1)

The space complexity consists of:

• cross matrix: 10 × 10 = 100 space, which is O(1) constant space
• vis array: size 10, which is O(1) constant space
• Recursion stack: maximum depth is n, which is at most 9, so O(1) constant space
• Local variables in each recursive call: O(1) per call

Since all space requirements are bounded by constants independent of input size, the overall space complexity is O(1).

Common Pitfalls

1. Incorrectly Handling the Skip Logic

Pitfall: A common mistake is misunderstanding when a middle dot needs to be visited. Developers often assume that ANY move between non-adjacent dots requires visiting an intermediate dot, but this is only true when the line passes directly through another dot's center.

Example of the mistake:

# WRONG: Assuming 2→9 requires visiting 5 first
if abs(current - next) > 1:  # This logic is too simplistic
    # Check for middle dot...

Why it's wrong: The move 2→9 doesn't require visiting 5 first because the line doesn't pass through dot 5's center. Only specific pairs like 1→3, 1→9, etc., have this restriction.

Solution: Use an explicit mapping table that only includes the pairs that actually require a middle dot:

skip_map = [[0] * 10 for _ in range(10)]
# Only set values for moves that actually cross through another dot's center
skip_map[1][3] = skip_map[3][1] = 2  # Horizontal through middle
skip_map[1][9] = skip_map[9][1] = 5  # Diagonal through center
# ... other specific pairs

2. Forgetting to Backtrack Properly

Pitfall: Failing to reset the visited state after exploring a path, which causes dots to remain marked as visited for subsequent explorations.

Example of the mistake:

def dfs(current, length):
    visited[current] = True
    result = 0
    for next_pos in range(1, 10):
        if not visited[next_pos]:
            result += dfs(next_pos, length + 1)
    # WRONG: Forgot to reset visited[current] = False
    return result

Why it's wrong: Without resetting, once a dot is visited in any path, it remains marked as visited for all subsequent paths, drastically reducing the count of valid patterns.

Solution: Always reset the visited state before returning:

def dfs(current, length):
    visited[current] = True
    result = 0
    # ... exploration logic ...
    visited[current] = False  # Critical backtracking step
    return result

3. Incorrect Base Case for Counting Valid Patterns

Pitfall: Only counting patterns when length equals exactly m or n, instead of counting all patterns with length in the range [m, n].

Example of the mistake:

# WRONG: Only counts patterns of exact lengths
if pattern_length == m or pattern_length == n:
    valid_patterns = 1
else:
    valid_patterns = 0

Why it's wrong: The problem asks for patterns with "at least m dots and at most n dots", meaning any length from m to n inclusive should be counted.

Solution: Count any pattern with length in the valid range:

# Correct: Count if length is anywhere in [m, n]
valid_patterns = 1 if pattern_length >= m else 0
# Continue exploring until pattern_length > n

4. Missing Symmetry Optimization or Applying It Incorrectly

Pitfall: Either not using symmetry at all (causing unnecessary computation) or incorrectly grouping symmetric positions.

Example of the mistake:

# WRONG: Treating all positions as unique
total = 0
for start in range(1, 10):
    total += dfs(start)

Why it's wrong: This calculates patterns for all 9 starting positions independently, missing the 4-fold symmetry of corners and edges.

Solution: Group symmetric positions correctly:

# Corners (1,3,7,9) are 4-fold symmetric
# Edges (2,4,6,8) are 4-fold symmetric  
# Center (5) is unique
return dfs(1) * 4 + dfs(2) * 4 + dfs(5)

5. Off-by-One Errors with Array Indexing

Pitfall: Using 0-based indexing for the dots when the problem naturally uses 1-9 numbering.

Example of the mistake:

# WRONG: Using 0-8 indexing
visited = [False] * 9
for next_pos in range(9):  # 0 to 8
    if not visited[next_pos]:
        # ...

Why it's wrong: This misaligns with the natural 1-9 numbering of the lock screen, making the skip_map logic confusing and error-prone.

Solution: Use 1-based indexing with size 10 arrays:

visited = [False] * 10  # Index 0 unused, indices 1-9 for dots
for next_pos in range(1, 10):  # 1 to 9
    if not visited[next_pos]:
        # ...