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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.
1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6 int longestIncreasingPath(vector<vector<int>>& matrix) {
7 if (matrix.empty() || matrix[0].empty()) return 0;
8 int m = matrix.size(), n = matrix[0].size();
9 vector<vector<int>> memo(m, vector<int>(n, -1));
10 int maxLen = 0;
11
12 std::function<int(int, int)> dfs = [&](int x, int y) {
13 if (memo[x][y] != -1) return memo[x][y];
14 vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
15 int max_len = 1;
16 for (const auto& d : directions) {
17 int nx = x + d.first, ny = y + d.second;
18 if (nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[nx][ny] > matrix[x][y]) {
19 max_len = max(max_len, 1 + dfs(nx, ny));
20 }
21 }
22 memo[x][y] = max_len;
23 return max_len;
24 };
25
26 for (int i = 0; i < m; ++i) {
27 for (int j = 0; j < n; ++j) {
28 maxLen = max(maxLen, dfs(i, j));
29 }
30 }
31 return maxLen;
32 }
33};In this C++ solution, we utilize a lambda function for the DFS which captures the variables by reference and uses the memoization technique to store results of previously computed paths. This approach efficiently calculates the longest increasing path starting from each cell and stores the result in the memo table to prevent redundant calculations.
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.
1public class Solution {
2 public int LongestIncreasingPath(int[][] matrix) {
3 if (matrix == null || matrix.Length == 0) return 0;
4 int m = matrix.Length, n = matrix[0].Length;
int[][] dp = new int[m][];
for (int i = 0; i < m; i++) dp[i] = new int[n];
int maxLen = 0;
int dfs(int x, int y) {
if (dp[x][y] != 0) return dp[x][y];
int[][] directions = new int[][]{new int[]{-1, 0}, new int[]{1, 0}, new int[]{0, -1}, new int[]{0, 1}};
dp[x][y] = 1;
foreach (var dir in directions) {
int nx = x + dir[0], ny = y + dir[1];
if (nx >= 0 && nx < m && ny >= 0 && ny < n && matrix[nx][ny] > matrix[x][y]) {
dp[x][y] = Math.Max(dp[x][y], 1 + dfs(nx, ny));
}
}
return dp[x][y];
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
maxLen = Math.Max(maxLen, dfs(i, j));
}
}
return maxLen;
}
}This C# solution follows a similar logic by using dynamic programming principles. The DFS function explores potential paths from each cell, leveraging an array dp to store already calculated results and determine the longest increasing path starting from a specific point.