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This approach uses a combination of Depth First Search (DFS) and memoization to solve the problem efficiently. We explore each cell, attempting to find the longest increasing path starting from that cell. To avoid recalculating the path length for each call from the same cell, we use memoization to store already computed results.
The time complexity is O(m * n) because each cell is computed once and cached. The space complexity is also O(m * n) due to the memoization table.
1class Solution {
2 public int longestIncreasingPath(int[][] matrix) {
3 if (matrix.length == 0) return 0;
4 int m = matrix.length, n = matrix[0].length;
5 int[][] memo = new int[m][n];
6 int maxLen = 0;
7 for (int i = 0; i < m; i++) {
8 for (int j = 0; j < n; j++) {
9 maxLen = Math.max(maxLen, dfs(matrix, i, j, memo));
10 }
11 }
12 return maxLen;
13 }
14
15 private int dfs(int[][] matrix, int x, int y, int[][] memo) {
16 if (memo[x][y] != 0) return memo[x][y];
17 int m = matrix.length, n = matrix[0].length;
18 int maxLen = 1;
19 int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
20 for (int[] d : directions) {
21 int nx = x + d[0], ny = y + d[1];
22 if (nx >= 0 && ny >= 0 && nx < m && ny < n && matrix[nx][ny] > matrix[x][y]) {
23 int len = 1 + dfs(matrix, nx, ny, memo);
24 maxLen = Math.max(maxLen, len);
25 }
26 }
27 memo[x][y] = maxLen;
28 return maxLen;
29 }
30}This Java solution uses a recursive DFS method dfs to explore all possible increasing paths from the current cell, storing intermediate results in a memoization table to avoid redundant calculations. Each cell's path length is calculated and cached, ensuring an efficient solution.
This approach systematically calculates the longest increasing path dynamically by first iterating over the matrix and then updating results based on previously computed values. By using a dynamic programming table, we determine, for each cell, the longest chain it can contribute to incrementally.
The time complexity is O(m * n) due to the fact that each matrix cell can be a starting point and is only calculated once with memoization guidance. The space complexity is O(m * n) because a result is stored for each cell in the DP table.
1var longestIncreasingPath = function(matrix)
The JavaScript solution uses dynamic programming. For each cell, the function dfs recursively explores valid neighbors, resulting in the longest increasing path starting from that cell. The result for each cell is stored in a DP table before being returned, which ensures efficiency by preventing the recalculation of already known results.