💡 Problem Formulation: Finding the longest increasing path in a matrix involves identifying the longest sequence of increasing values that one can traverse in the matrix, moving only up, down, left, or right. For instance, given a matrix such as [[9,9,4],[6,6,8],[2,1,1]], the longest increasing path is [1, 2, 6, 9], resulting in an output length of 4.
Method 1: Depth-First Search (DFS) with Memoization
The depth-first search algorithm explores as far as possible down each branch before backtracking. When combined with memoization, which stores the results of expensive function calls, we optimize the computation by avoiding repeated calculations for the same inputs. This method is well-suited for the longest increasing path problem, as it greatly reduces the search space and time complexity.
Here’s an example:
def dfs(matrix, i, j, memo):
if memo[i][j]:
return memo[i][j]
directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
for d in directions:
x, y = i + d[0], j + d[1]
if 0 <= x < len(matrix) and 0 <= y matrix[i][j]:
memo[i][j] = max(memo[i][j], dfs(matrix, x, y, memo))
memo[i][j] += 1
return memo[i][j]
def longest_increasing_path(matrix):
if not matrix or not matrix[0]:
return 0
rows, cols = len(matrix), len(matrix[0])
memo = [[0]*cols for _ in range(rows)]
return max(dfs(matrix, i, j, memo) for i in range(rows) for j in range(cols))
Output:
4
This function, longest_increasing_path(matrix), initializes a memoization matrix and iteratively applies the DFS to each element. The dfs() function itself returns the length of the longest increasing path starting from the given cell. Memoization ensures that each cell is processed only once, making the solution highly efficient.
Method 2: Dynamic Programming
Dynamic programming solves problems by combining the solutions of subproblems. In the context of finding the longest increasing path, it entails build…