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Longest Increasing Path in a Matrix⚓︎

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Description⚓︎

Given an \(m\times n\) 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] <= 2^31 - 1

Solution⚓︎

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class Solution {
private:
    int n, m;
    vector<vector<int>> f;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

    int dp(int x, int y, vector<vector<int>>& matrix) {
        if (f[x][y] != -1) return f[x][y];

        f[x][y] = 1;
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 0 && a < n && b >= 0 && b < m && matrix[a][b] > matrix[x][y])
                f[x][y] = max(f[x][y], dp(a, b, matrix) + 1);
        }

        return f[x][y];
    }
public:
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        n = matrix.size(), m = matrix[0].size();
        f = vector<vector<int>>(n, vector<int>(m, -1));

        int res = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++)
                res = max(res, dp(i, j, matrix));
        }
        return res;
    }
};