How do you solve Longest Increasing Path in a Matrix on LeetCode?

Longest Increasing Path in a Matrix (LeetCode #329) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: The longest increasing path can start from any cell.

What is the brute force solution for Longest Increasing Path in a Matrix?

The brute force approach for Longest Increasing Path in a Matrix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible paths from each cell to find the longest increasing path. This means we will recursively explore all four directions from each cell, keeping track of the length of the path. The optimal Optimal Solution approach improves this to O(m * n).

What are the interview tips for Longest Increasing Path in a Matrix?

Explain your thought process clearly while coding. Discuss potential optimizations as you write your initial solution. Practice dry-running your code with sample inputs before finalizing.

What are common mistakes when solving Longest Increasing Path in a Matrix?

Not using memoization, leading to excessive recursive calls. Forgetting to check boundaries when exploring neighbors.

#329

Longest Increasing Path in a Matrix

Hard

ArrayDynamic ProgrammingDepth-First SearchBreadth-First SearchGraph TheoryTopological SortMemoizationMatrixDepth-First SearchMemoizationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses memoization to store the results of previously computed paths, significantly reducing redundant calculations. This allows us to efficiently find the longest increasing path without re-exploring paths we've already computed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a memoization table to store the length of the longest increasing path starting from each cell.
2Step 2: For each cell, if the result is not already computed, perform a DFS to compute the longest path, storing results in the memo table.
3Step 3: Return the maximum value found in the memo table.

solution.py24 lines

1# Full working Python code
2class Solution:
3    def longestIncreasingPath(self, matrix):
4        if not matrix:
5            return 0
6        m, n = len(matrix), len(matrix[0])
7        self.memo = [[-1] * n for _ in range(m)]
8        max_length = 0
9        for i in range(m):
10            for j in range(n):
11                max_length = max(max_length, self.dfs(matrix, i, j))
12        return max_length
13
14    def dfs(self, matrix, x, y):
15        if self.memo[x][y] != -1:
16            return self.memo[x][y]
17        max_length = 1
18        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
19        for dx, dy in directions:
20            nx, ny = x + dx, y + dy
21            if 0 <= nx < len(matrix) and 0 <= ny < len(matrix[0]) and matrix[nx][ny] > matrix[x][y]:
22                max_length = max(max_length, 1 + self.dfs(matrix, nx, ny))
23        self.memo[x][y] = max_length
24        return max_length

ℹ

Complexity note: The time complexity is O(m * n) because we compute the longest path for each cell exactly once, thanks to memoization. The space complexity is O(m * n) due to the memoization table storing results for each cell.

1The longest increasing path can start from any cell.
2Memoization is crucial to avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.