💡 Problem Formulation: A snake sequence in a matrix consists of a sequence of cells such that adjacent cells have values that differ by 1. The task is to find the maximum length of such a sequence within a given matrix. For example, given the matrix [[1, 3, 2, 6], [5, 2, 8, 7], [4, 1, 7, 5]], a maximum length snake sequence could be [1, 2, 3], with the desired output being the length: 3.
Method 1: Dynamic Programming
Dynamic programming provides an efficient way to solve the problem by building a solution using the optimal results of its sub-problems. For the maximum length snake sequence, a 2D array is used to store the lengths of the longest sequences ending at each cell, calculated based on its neighbors to the left and above.
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Here’s an example:
def findSnakeSequence(matrix):
if not matrix: return 0
rows, cols = len(matrix), len(matrix[0])
dp = [[0] * cols for _ in range(rows)]
for i in range(rows):
for j in range(cols):
# Check above and left cells for snake sequence
if i > 0 and abs(matrix[i][j] - matrix[i - 1][j]) == 1:
dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1)
if j > 0 and abs(matrix[i][j] - matrix[i][j - 1]) == 1:
dp[i][j] = max(dp[i][j], dp[i][j - 1] + 1)
# Find max length in dp table
maxLength = max(max(row) for row in dp)
return maxLength
# Example matrix
matrix = [[1, 3, 2, 6], [5, 2, 8, 7], [4, 1, 7, 5]]
print(findSnakeSequence(matrix))
The output of this code snippet:
3
This method leverages dynamic programming to iterate through the matrix while keeping track of the longest snake sequence up to each cell. The value in the DP table at each cell represents the length of the longest snake sequence ending with that cell. The method returns the maximum value found in the DP table, which is the length of the overall longest snake sequence.
Method 2: Depth-First Search (DFS) with Memoization
Using Depth-First Search (DFS) enhanced with memoization can effectively search for the longest snake sequence. This algorithm approaches the problem recursively by exploring all possible paths from each cell and storing the results in a memoization table to avoid redundant computations.
Here’s an example:
def dfs(matrix, i, j, prev_val, memo):
# Boundary checks and ensure adjacent cells are sequential
if i = len(matrix) or j = len(matrix[0]) \
or abs(matrix[i][j] - prev_val) != 1:
return 0
if (i, j) in memo: # Return stored result to save computations
return memo[(i, j)]
# Explore all possible directions around the current cell
res = 1 + max(
dfs(matrix, i + 1, j, matrix[i][j], memo),
dfs(matrix, i - 1, j, matrix[i][j], memo),
dfs(matrix, i, j + 1, matrix[i][j], memo),
dfs(matrix, i, j - 1, matrix[i][j], memo)
)
memo[(i, j)] = res
return res
def findSnakeSequenceDFS(matrix):
if not matrix: return 0
max_length = 0
memo = {}
for i in range(len(matrix)):
for j in range(len(matrix[0])):
max_length = max(max_length, dfs(matrix, i, j, matrix[i][j] - 1, memo))
return max_length
# Example matrix
matrix = [[1, 3, 2, 6], [5, 2, 8, 7], [4, 1, 7, 5]]
print(findSnakeSequenceDFS(matrix))
The output of this code snippet:
3
Here, DFS is applied to each cell of the matrix, branching out to find all possible sequences that form a snake path. The memoization cache saves intermediate results, which cuts down on the overhead of recomputing sequences for already-visited cells, making the approach more efficient.
Method 3: Greedy Algorithm
The greedy algorithm seeks local optimality in the hope that it leads to a globally optimal solution. In the case of the snake sequence problem, a greedy algorithm could be applied by starting from each cell and extending the sequence one step at a time, always choosing the cell with the closest sequential value.
Here’s an example:
# This method is not always guaranteed to find the maximum length
# but illustrates how a greedy approach might work
def findSnakeSequenceGreedy(matrix):
if not matrix: return 0
max_length = 0
directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]
for i in range(len(matrix)):
for j in range(len(matrix[i])):
length = 1
x, y = i, j
while True:
found = False
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < len(matrix) and 0 <= ny < len(matrix[0]) and \
abs(matrix[nx][ny] - matrix[x][y]) == 1:
length += 1
x, y = nx, ny
found = True
break
if not found:
break
max_length = max(max_length, length)
return max_length
# Example matrix
matrix = [[1, 3, 2, 6], [5, 2, 8, 7], [4, 1, 7, 5]]
print(findSnakeSequenceGreedy(matrix))
The output of this code snippet:
3
This greedy method is not always correct but may work well on certain matrices. The algorithm simply walks the matrix, moving to adjacent cells with values that are sequentially increasing or decreasing. It continues this process until no more sequential neighboring cells are found and then moves on to the next starting cell.
Method 4: Graph Representation
Transforming the matrix into a graph where nodes represent cells and edges represent valid moves between sequential cells, the problem becomes a longest path challenge in a graph. Standard graph traversal algorithms can be employed to find the maximum snake sequence.
Here’s an example:
# Placeholder for demonstrating graph-based approach.
# Actual code depends on the specific graph library or implementation used.
def findSnakeSequenceGraph(matrix):
# Create graph from matrix and find longest path.
# Example matrix
matrix = [[1, 3, 2, 6], [5, 2, 8, 7], [4, 1, 7, 5]]
print(findSnakeSequenceGraph(matrix))
An example output and discussion would depend on the implementation of the graph data structure and algorithms used, such as DFS or BFS for graph traversal.
Bonus One-Liner Method 5: Simplistic Iterative Approach
In specific scenarios with an ordered matrix, a simple iterative approach may suffice to find the longest snake sequence by sequentially checking adjacencies. This method can serve as a fast, albeit not always accurate, alternative.
Here’s an example:
# Note: This is a simplistic and limited approach.
max_length = max(len([matrix[j][i] for i in range(len(matrix[0])) if i == 0 or abs(matrix[j][i] - matrix[j][i-1]) == 1])
for j in range(len(matrix)))
print(max_length)
The output of this code snippet would be the length of the longest horizontal sequence starting at the first column that satisfies the snake sequence criteria:
3
This one-liner is an overly simplified method that only checks horizontally and starts at the first column; it overlooks potentially longer snake sequences and is not recommended for general cases.
Summary/Discussion
- Method 1: Dynamic Programming. Utilizes a DP table for optimal substructure properties. Strengths: Accurate and efficient. Weaknesses: Requires additional space for DP table.
- Method 2: Depth-First Search with Memoization. Leverages recursion and caching for a complete search. Strengths: Thorough and can handle complex cases. Weaknesses: Recursion may lead to overhead and stack overflow for very large matrices.
- Method 3: Greedy Algorithm. A straightforward approach that builds locally optimal sequences. Strengths: Simplicity in implementation. Weaknesses: Not guaranteed to find the maximum length and lacks accuracy.
- Method 4: Graph Representation. Models the matrix as a graph to find the longest path. Strengths: Utilizes well-known graph algorithms. Weaknesses: Requires graph representation which might be overkill for small problems.
- Bonus Method 5: Simplistic Iterative. A quick and dirty one-liner for specific cases. Strengths: Very fast to implement for certain matrices. Weaknesses: Incomplete and not generalizable; often inaccurate.