329. Longest Increasing Path in a Matrix
Description
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).
Constraints
m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
0 <= matrix[i][j] <= 231 - 1
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
Input: matrix = [[9, 9, 4], [6, 6, 8], [2, 1, 1]]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].

Solutions
/**
* Time complexity :
* Space complexity :
*/
class Solution {
private final int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
public int longestIncreasingPath(int[][] matrix) {
int result = 0;
if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return result;
}
int rows = matrix.length;
int cols = matrix[0].length;
int[][] mem = new int[rows][cols];
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
result = Math.max(result, longestIncreasingPath(matrix, mem, i, j));
}
}
return result;
}
private int longestIncreasingPath(int[][] matrix, int[][] mem, int i, int j) {
if(mem[i][j] > 0) {
return mem[i][j];
}
for(int k = 0; k < dir.length; k++) {
int x = i + dir[k][0];
int y = j + dir[k][1];
if(x >= 0 && x < matrix.length && y >= 0 && y < matrix[0].length && matrix[x][y] > matrix[i][j]) {
mem[i][j] = Math.max(mem[i][j], longestIncreasingPath(matrix, mem, x, y));
}
}
return ++mem[i][j];
}
}
Follow up
Last updated
Was this helpful?