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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:
[1, 2, 6, 9]
Example 2:
[3, 4, 5, 6]
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
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 $\max_{i, j} \textit{dfs}(i, j)$.
The execution logic of the function $dfs(i, j)$ is as follows:
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.
Similar problems:
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))
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];
}
}
class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
int f[m][n];
memset(f, 0, sizeof(f));
int ans = 0;
int dirs[5] = {-1, 0, 1, 0, -1};
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (f[i][j]) {
return f[i][j];
}
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
f[i][j] = max(f[i][j], dfs(x, y));
}
}
return ++f[i][j];
};
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
ans = max(ans, dfs(i, j));
}
}
return ans;
}
};
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
}
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;
}
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