LeetCode #329 — HARD

Longest Increasing Path in a Matrix

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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, 4, 5, 6]. Moving diagonally is not allowed.

Example 3:

Input: matrix = [[1]]
Output: 1

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Topological Sort

Example 1

[[9,9,4],[6,6,8],[2,1,1]]

Example 2

[[3,4,5],[3,2,6],[2,2,1]]

Example 3

[[1]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #329: Longest Increasing Path in a Matrix
class Solution {
    private int m;
    private int n;
    private int[][] matrix;
    private int[][] f;

    public int longestIncreasingPath(int[][] matrix) {
        m = matrix.length;
        n = matrix[0].length;
        f = new int[m][n];
        this.matrix = matrix;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (f[i][j] != 0) {
            return f[i][j];
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                f[i][j] = Math.max(f[i][j], dfs(x, y));
            }
        }
        return ++f[i][j];
    }
}

// Accepted solution for LeetCode #329: Longest Increasing Path in a Matrix
func longestIncreasingPath(matrix [][]int) (ans int) {
	m, n := len(matrix), len(matrix[0])
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if f[i][j] != 0 {
			return f[i][j]
		}
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if 0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j] {
				f[i][j] = max(f[i][j], dfs(x, y))
			}
		}
		f[i][j]++
		return f[i][j]
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			ans = max(ans, dfs(i, j))
		}
	}
	return
}

# Accepted solution for LeetCode #329: Longest Increasing Path in a Matrix
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))

// Accepted solution for LeetCode #329: Longest Increasing Path in a Matrix
impl Solution {
    pub fn longest_increasing_path(matrix: Vec<Vec<i32>>) -> i32 {
        use std::cmp::max;
        use std::collections::HashMap;

        let mut dp: HashMap::<(usize, usize), i32> = HashMap::new();

        fn dfs(
            r: usize,
            c: usize,
            prevVal: i32,
            matrix: &[Vec<i32>],
            dp: &mut HashMap<(usize, usize), i32>,
        ) -> i32 {
            if r < 0
                || r >= matrix.len()
                || c < 0
                || c >= matrix[0].len()
                || matrix[r][c] <= prevVal
            {
                return 0;
            }
            if let Some(&result) = dp.get(&(r, c)) {
                return result;
            }

            let mut res = 1;
            res = max(res, 1 + dfs(r + 1, c, matrix[r][c], matrix, dp));
            res = max(res, 1 + dfs(r - 1, c, matrix[r][c], matrix, dp));
            res = max(res, 1 + dfs(r, c + 1, matrix[r][c], matrix, dp));
            res = max(res, 1 + dfs(r, c - 1, matrix[r][c], matrix, dp));
            dp.insert((r, c), res);
            res
        }

        for r in 0..matrix.len() {
            for c in 0..matrix[0].len() {
                dfs(r, c, -1, &matrix, &mut dp);
            }
        }
        *dp.values().max().unwrap_or(&1)
    }
}

// Accepted solution for LeetCode #329: Longest Increasing Path in a Matrix
function longestIncreasingPath(matrix: number[][]): number {
    const m = matrix.length;
    const n = matrix[0].length;
    const f: number[][] = Array(m)
        .fill(0)
        .map(() => Array(n).fill(0));
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        if (f[i][j] > 0) {
            return f[i][j];
        }
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
                f[i][j] = Math.max(f[i][j], dfs(x, y));
            }
        }
        return ++f[i][j];
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans = Math.max(ans, dfs(i, j));
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.