#329

Hard

auto_awesomeGraph indegree plus topological ordering

LeetCode Problem Workspace

Longest Increasing Path in a Matrix

Find the length of the longest increasing path in a matrix with given movement constraints using graph techniques.

array dynamic programming depth first search breadth first search graph

Problem Statement

You are given an m x n matrix filled with integers, where each element represents a value. The task is to return the length of the longest strictly increasing path in the matrix. From any cell, you may move in four directions: up, down, left, or right. You cannot move diagonally or outside the boundaries of the matrix.

The problem challenges you to efficiently compute this path, ensuring that each movement leads to an increasing value. The matrix size is up to 200x200, and values can be as large as 2^31 - 1. The solution requires handling large input sizes within a reasonable time frame.

Examples

Example 1

Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]

Output: 4

The longest increasing path is [1, 2, 6, 9].

Example 2

Input: matrix = [[3,4,5],[3,2,6],[2,2,1]]

Output: 4

The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Example 3

Input: matrix = [[1]]

Output: 1

Example details omitted.

Constraints

m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
0 <= matrix[i][j] <= 231 - 1

Solution Approach

Graph Indegree and Topological Sort

The problem can be modeled as a graph where each cell is a node and edges represent possible movements to adjacent cells with greater values. By calculating the indegree (number of incoming edges) for each node, we can use topological sorting to determine the longest increasing path in linear time.

Depth-First Search (DFS) with Memoization

DFS is used to explore the longest path from each cell, while memoization stores previously computed results to avoid redundant calculations. This optimizes performance, especially for large matrices, by ensuring that each cell's result is calculated once.

Breadth-First Search (BFS) for Level-wise Processing

BFS can also be applied by processing the matrix in a level-order manner, starting from the cells with the smallest values. This approach is useful when the matrix structure suggests processing in layers based on value comparisons.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depends on the chosen approach. For the DFS approach with memoization, the time complexity is O(m * n), where m and n are the matrix dimensions. Space complexity also scales with the matrix size due to the storage of memoization results. Topological sorting with graph indegree typically yields similar complexities.

What Interviewers Usually Probe

The candidate should understand how to efficiently traverse the matrix using graph-like techniques such as topological sort.
Look for a clear explanation of how DFS with memoization avoids recalculating paths.
Expect discussion on how to handle large input sizes and edge cases such as matrices with only one cell or uniform values.

Common Pitfalls or Variants

Common pitfalls

Ignoring boundary conditions and invalid moves (e.g., out-of-bound or non-increasing paths).
Failing to optimize with memoization, leading to redundant DFS calls and time inefficiency.
Overcomplicating the problem by using unnecessary algorithms instead of focusing on graph-based approaches.

Follow-up variants

Handling a matrix with only one cell.
Optimizing for larger matrix sizes by reducing the overhead of DFS or BFS.
Implementing the solution without using memoization or dynamic programming techniques.

FAQ

How do I use graph indegree for this problem?

Graph indegree helps determine the sequence in which cells should be processed. By sorting the cells based on their indegree, we ensure that each cell's path is calculated only after all possible moves to greater values have been considered.

What is the best way to optimize the solution?

Using DFS with memoization is the best way to optimize the solution. This ensures that each cell's longest path is computed only once, avoiding redundant calculations and improving performance.

What is the time complexity of this problem?

The time complexity is O(m * n) for DFS with memoization, where m is the number of rows and n is the number of columns in the matrix. This is because each cell is processed once.

Can BFS be used for solving this problem?

Yes, BFS can be applied in this problem by processing cells in layers based on value comparisons. While BFS might not be the most common approach, it is still valid.

How does GhostInterview help with solving matrix traversal problems?

GhostInterview offers structured solutions and insights into optimizing matrix traversal problems, such as applying DFS with memoization and leveraging graph algorithms.

terminal

Solution

Solution 1: Memoization Search

We design a function $dfs(i, j)$, which represents the length of the longest increasing path that can be obtained starting from the coordinate $(i, j)$ in the matrix. The answer is $\max_{i, j} \textit{dfs}(i, j)$.

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class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            ans = 0
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]:
                    ans = max(ans, dfs(x, y))
            return ans + 1

        m, n = len(matrix), len(matrix[0])
        return max(dfs(i, j) for i in range(m) for j in range(n))

Continue Practicing

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