What is the time complexity of Longest Increasing Path in a Matrix?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Longest Increasing Path in a Matrix?

The space complexity is O(m*n).

What data structures are used to solve Longest Increasing Path in a Matrix?

This problem uses Array, Dynamic Programming, Depth-First Search, Breadth-First Search, Graph, Topological Sort, Memoization, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Longest Increasing Path in a Matrix?

Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths starting from each cell. Insight 2: Memoization is crucial to avoid redundant calculations. Since the longest increasing path starting from a cell (r, c) is the same regardless of how we reached that cell, we can store the length in a dp array to prevent recomputation.

Which companies ask LeetCode 329. Longest Increasing Path in a Matrix in interviews?

This problem is frequently asked at Citadel, DE Shaw, DoorDash, Nvidia, Snap and 1 more companies.

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Longest Increasing Path in a Matrix - LeetCode 329 Solution

Longest Increasing Path in a Matrix - Complete Solution Guide

Longest Increasing Path in a Matrix is LeetCode problem 329, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an m x n integers matrix , return the length of the longest increasing path in matrix . From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed). Example 1: Input: matrix = [[9,9,4],[6,6,8],[2,1,1]] Output: 4 Explanation: The longest increasing path is [1, 2, 6, 9] . Example 2: Input: matrix = [[3,4,5],[3,2,6],[2,2,1]] Output: 4 Explanation: The longest increasing path is [3

Detailed Explanation

The problem asks us to find the length of the longest increasing path in a given matrix of integers. We can move from a cell to its adjacent cells (up, down, left, or right) if the value of the adjacent cell is strictly greater than the current cell. We cannot move diagonally or wrap around the boundaries of the matrix. The goal is to find a path of cells where each cell's value is greater than the previous cell, and this path is the longest possible.

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm with memoization (dynamic programming). We iterate through each cell of the matrix and initiate a DFS search starting from that cell. The DFS function explores all possible increasing paths from the current cell to its neighbors. To avoid recomputing the same path lengths, we use a `dp` array to store the length of the longest increasing path starting from each cell. The base case for the DFS is when the length of the longest increasing path from a cell has already been computed (i.e., `dp[r][c] > 0`). We also check for boundary conditions and the increasing path condition when exploring neighbors. The algorithm returns the maximum length of the longest increasing path found among all starting cells.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of the same dimensions as the input matrix, filled with zeros. This array will store the length of the longest increasing path starting from each cell.
Step 2: Iterate through each cell (r, c) in the matrix.
Step 3: For each cell, call a DFS function `dfs(r, c)` to find the length of the longest increasing path starting from that cell.
Step 4: The `dfs(r, c)` function first checks if the value in `dp[r][c]` is greater than 0. If it is, it means the length of the longest increasing path starting from this cell has already been computed, so it returns `dp[r][c]` directly.
Step 5: If `dp[r][c]` is 0, it means the length hasn't been computed yet. The function then explores the four neighbors of the cell (up, down, left, right).
Step 6: For each neighbor (nr, nc), it checks if the neighbor is within the boundaries of the matrix and if the value of the neighbor is strictly greater than the value of the current cell (matrix[nr][nc] > matrix[r][c]).
Step 7: If both conditions are met, it recursively calls `dfs(nr, nc)` to find the length of the longest increasing path starting from the neighbor.
Step 8: The `dfs` function keeps track of the maximum length of the longest increasing path found among all neighbors and adds 1 to it (to account for the current cell).
Step 9: The computed length is stored in `dp[r][c]`, and the function returns this length.
Step 10: After iterating through all cells in the matrix, the algorithm returns the maximum value found in the `dp` array, which represents the length of the longest increasing path in the matrix.

Key Insights

Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths starting from each cell.
Insight 2: Memoization is crucial to avoid redundant calculations. Since the longest increasing path starting from a cell (r, c) is the same regardless of how we reached that cell, we can store the length in a dp array to prevent recomputation.
Insight 3: The increasing path condition requires checking neighbors' values, and we must ensure we don't move to a cell with a smaller or equal value. The boundary conditions also need to be checked carefully.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Depth-First Search, Breadth-First Search, Graph, Topological Sort, Memoization, Matrix.

Companies

Asked at: Citadel, DE Shaw, DoorDash, Nvidia, Snap, WeRide.

Longest Increasing Path in a Matrix | LeetCode 329

Hard

Array Dynamic Programming Depth-First Search Breadth-First Search Graph Topological Sort Memoization Matrix

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

Solution Code

Available in:

class Solution {
    public int longestIncreasingPath(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }

        int m = matrix.length;
        int n = matrix[0].length;
        int[][] dp = new int[m][n];

        int max_len = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                max_len = Math.max(max_len, dfs(matrix, dp, i, j));
            }
        }

        return max_len;
    }

    private int dfs(int[][] matrix, int[][] dp, int r, int c) {
        if (dp[r][c] > 0) {
            return dp[r][c];
        }

        int m = matrix.length;
        int n = matrix[0].length;
        int max_from_neighbors = 0;
        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        for (int[] dir : directions) {
            int nr = r + dir[0];
            int nc = c + dir[1];

            if (0 <= nr && nr < m && 0 <= nc && nc < n && matrix[nr][nc] > matrix[r][c]) {
                max_from_neighbors = Math.max(max_from_neighbors, dfs(matrix, dp, nr, nc));
            }
        }

        dp[r][c] = 1 + max_from_neighbors;
        return dp[r][c];
    }
}

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Problem Info

Description

Problem 329: Longest Increasing Path in a Matrix - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Depth-First Search Breadth-First Search Graph Topological Sort Memoization Matrix

Companies

CitadelDE ShawDoorDashNvidiaSnapWeRide

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:329

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a `dp` array of the same dimensions as the input matrix, filled with zeros. This array will store the length of the longest increasing path starting from each cell.
2
Step 2: Iterate through each cell (r, c) in the matrix.
3
Step 3: For each cell, call a DFS function `dfs(r, c)` to find the length of the longest increasing path starting from that cell.
4
Step 4: The `dfs(r, c)` function first checks if the value in `dp[r][c]` is greater than 0. If it is, it means the length of the longest increasing path starting from this cell has already been computed, so it returns `dp[r][c]` directly.
5
Step 5: If `dp[r][c]` is 0, it means the length hasn't been computed yet. The function then explores the four neighbors of the cell (up, down, left, right).
6
Step 6: For each neighbor (nr, nc), it checks if the neighbor is within the boundaries of the matrix and if the value of the neighbor is strictly greater than the value of the current cell (matrix[nr][nc] > matrix[r][c]).
7
Step 7: If both conditions are met, it recursively calls `dfs(nr, nc)` to find the length of the longest increasing path starting from the neighbor.
8
Step 8: The `dfs` function keeps track of the maximum length of the longest increasing path found among all neighbors and adds 1 to it (to account for the current cell).
9
Step 9: The computed length is stored in `dp[r][c]`, and the function returns this length.
10
Step 10: After iterating through all cells in the matrix, the algorithm returns the maximum value found in the `dp` array, which represents the length of the longest increasing path in the matrix.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

Time Complexity: Each cell in the matrix is visited at most once by the DFS function due to memoization. For each cell, we perform a constant amount of work (checking its neighbors). Therefore, the total time complexity is O(m*n), where m is the number of rows and n is the number of columns in the matrix. Space Complexity: The space complexity is dominated by the `dp` array, which has dimensions m x n. The recursive call stack of the DFS can also contribute to the space complexity, but in the worst case, the depth of the recursion is limited by the longest increasing path, which is at most m*n. Therefore, the overall space complexity is O(m*n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Topological Sort

An alternative approach involves treating the matrix as a directed graph, where edges exist from cell A to cell B if A's value is smaller than B's and A and B are adjacent. Then, we can perform a topological sort to find the longest path. The length of the longest path will be the longest increasing path in the matrix.

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

Trade-offs: While also O(m*n) in terms of time and space, this approach can be more complex to implement. The DFS with memoization approach is often more intuitive and easier to code.

✨

Key Insights

1Insight 1: The problem can be solved using Depth-First Search (DFS) because we need to explore all possible paths starting from each cell.
2Insight 2: Memoization is crucial to avoid redundant calculations. Since the longest increasing path starting from a cell (r, c) is the same regardless of how we reached that cell, we can store the length in a dp array to prevent recomputation.
3Insight 3: The increasing path condition requires checking neighbors' values, and we must ensure we don't move to a cell with a smaller or equal value. The boundary conditions also need to be checked carefully.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: Empty matrix or matrix with zero rows/columns. The solution handles this by returning 0 at the beginning.
Edge case 2: Matrix with only one element. In this case, the algorithm will correctly identify that the longest increasing path is of length 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to memoize the results of DFS. Without memoization, the algorithm would have a much higher time complexity (exponential) and likely exceed the time limit for larger matrices.

✗Common mistake 2: Failing to check boundary conditions when exploring neighbors. This can lead to accessing invalid memory locations and causing runtime errors.

✗Common mistake 3: Incorrectly implementing the increasing path condition. It is important to ensure that the neighbor's value is strictly greater than the current cell's value.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in geographical data where the elevation represents the value. Finding the longest increasing path could represent the longest continuous climb in a terrain.

🏦Real-world scenario 2: Analyzing trends in financial data represented as a matrix. Identifying the longest increasing path can help determine the longest period of continuous growth in a specific sector or asset.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 28, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Longest Increasing Path in a Matrix"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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