329. Longest Increasing Path in a Matrix

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

Example: 2

Input: matrix = [[3,4,5],[3,2,6],[2,2,1]]

Output: 4

Example: 3

Input: matrix = [[1]]

Output: 1

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 200
• 0 <= matrix[i][j] <= 2 ^ 31 - 1

Step I (Making Observations):

• Important observations to make here is that, we can have atleast 1 as our answer, if we take only one block in path.
• We can only from current cell to its neighbouring cell if and only if we the neighbouring cell has greater value than current cell.
• Because of this we can see that there is only one way path, so the entire grid can be represented as a directed acyclic graph (DAG).
• This problem is similar to finding the longest path in a DAG.
• We need not create any graph explicitly here, we will solve it using recursion.

Step II (Solving the problem using Recursion):

• Now for each cell [i, j], we will find the longest path that can be taken if this current cell is considered.
• So the recursion goes this way: From current cell we will check whether we can go to any other neighbouring cell.
• If YES we will go to that cell, and if NO we will return 1 (because we are considering the current cell).
• This step is shown in code below.

Solving using recursion (Will give TLE): int recur(int r, int c, int[][] matrix) { int max = 0; for(int i = 0; i < 4; i++) { // dr = new int[] {1, 0, -1, 0} // dc = new int[] {0, 1, 0, -1} int R = r + dr[i]; int C = c + dc[i]; if(R < 0 || C < 0 || R >= matrix.length || C >= matrix[0].length) { continue; } // we have found a path... so go to that cell if(matrix[R][C] > matrix[r][c]) { max = Math.max(max, recur(R, C, matrix)); } } return max + 1; }

Step III (Using DP: From TLE to Accepted):

• Now for every state [i, j] we might be recomputing them again and again during recursion.
• This can result into TLE.
• So what we will do is, after computing value of [i, j] we will store it into dp array mapping them by [i, j].
• This step is shown in code below.

Full Code (Java): class Solution { int[] dr, dc, dp[]; public int longestIncreasingPath(int[][] matrix) { dr = new int[] {1, 0, -1, 0}; dc = new int[] {0, 1, 0, -1}; int n = matrix.length; int m = matrix[0].length; dp = new int[n][m]; int max = 1; for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { max = Math.max(max, recur(i, j, matrix)); } } return max; } int recur(int r, int c, int[][] matrix) { int max = 0; // checking if the result for this subproblem is stored // if yes we will return this value directly without further computation if(dp[r][c] != 0) { return dp[r][c]; } for(int i = 0; i < 4; i++) { int R = r + dr[i]; int C = c + dc[i]; if(R < 0 || C < 0 || R >= matrix.length || C >= matrix[0].length) { continue; } if(matrix[R][C] > matrix[r][c]) { max = Math.max(max, recur(R, C, matrix)); } } // storing the value of current subproblem in DP dp[r][c] = max + 1; return max + 1; } }