329

329. Longest Increasing Path in a Matrix (Hard)

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
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: matrix = [[3,4,5],[3,2,6],[2,2,1]]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

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

Related Topics:
Depth-first Search, Topological Sort, Memoization

Solution 1. Topological Sort (DFS)

A DFS version topoligical sort is Post-order Traversal + Memo.

Use a vector<vector<int>> cnt, where cnt[i][j] is the length of longest increasing path starting from matrix[i][j]. Initially values in cnt are all zeroes.

For each position matrix[x][y],

• if cnt[x][y] is not zero, which means it's already visited, return cnt[x][y] right away.
• otherwise, probe the 4 directions, cnt[x][y] is one greater than its largest neightbor.

// OJ: https://leetcode.com/problems/longest-increasing-path-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
    vector<vector<int>> cnt;
    int ans = 0, M, N, dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
    int dfs(vector<vector<int>> &A, int x, int y) {
        if (cnt[x][y]) return cnt[x][y];
        cnt[x][y] = 1;
        for (auto &dir : dirs) {
            int a = x + dir[0], b = y + dir[1];
            if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] <= A[x][y]) continue;
            cnt[x][y] = max(cnt[x][y], 1 + dfs(A, a, b));
        }
        return cnt[x][y];
    }
public:
    int longestIncreasingPath(vector<vector<int>>& A) {
        if (A.empty() || A[0].empty()) return 0;
        M = A.size(), N = A[0].size();
        cnt.assign(M, vector<int>(N));
        for (int i = 0; i < M; ++i) 
            for (int j = 0; j < N; ++j) 
                ans = max(ans, dfs(A, i, j));
        return ans;
    }
};

Solution 2. Topological Sort (BFS)

// OJ: https://leetcode.com/problems/longest-increasing-path-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int longestIncreasingPath(vector<vector<int>>& A) {
        int M = A.size(), N = A[0].size(), dirs[4][2] = {{0,1},{0,-1},{1,0},{-1,0}}, ans = 0;
        vector<vector<int>> indegree(M, vector<int>(N));
        queue<pair<int, int>> q;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                for (auto &[dx, dy] : dirs) {
                    int a = i + dx, b = j + dy;
                    if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] >= A[i][j]) continue;
                    indegree[i][j]++;
                }
                if (indegree[i][j] == 0) q.push({ i, j });
            }
        }
        while (q.size()) {
           int cnt = q.size(); 
            ++ans;
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] <= A[x][y]) continue;
                    if (--indegree[a][b] == 0) q.push({ a, b });
                }
            }
        }
        return ans;
    }
};