Question

329. Longest Increasing Path in a Matrix

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Description

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

Solutions

Solution 1: Memoization Search

We design a function $dfs(i, j)$, which represents the length of the longest increasing path that can be obtained starting from the coordinate $(i, j)$ in the matrix. The answer is $\max_{i, j} \textit{dfs}(i, j)$.

The execution logic of the function $dfs(i, j)$ is as follows:

• If $(i, j)$ has been visited, directly return $\textit{f}(i, j)$;
• Otherwise, search $(i, j)$, search the coordinates $(x, y)$ in four directions. If $0 \le x < m, 0 \le y < n$ and $matrix[x][y] > matrix[i][j]$, then search $(x, y)$. After the search is over, update $\textit{f}(i, j)$ to $\textit{f}(i, j) = \max(\textit{f}(i, j), \textit{f}(x, y) + 1)$. Finally, return $\textit{f}(i, j)$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

Similar problems:

• 2328. Number of Increasing Paths in a Grid

class Solution:
  def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
    @cache
    def dfs(i: int, j: int) -> int:
      ans = 0
      for a, b in pairwise((-1, 0, 1, 0, -1)):
        x, y = i + a, j + b
        if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]:
          ans = max(ans, dfs(x, y))
      return ans + 1

    m, n = len(matrix), len(matrix[0])
    return max(dfs(i, j) for i in range(m) for j in range(n))

class Solution {
  private int m;
  private int n;
  private int[][] matrix;
  private int[][] f;

  public int longestIncreasingPath(int[][] matrix) {
    m = matrix.length;
    n = matrix[0].length;
    f = new int[m][n];
    this.matrix = matrix;
    int ans = 0;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans = Math.max(ans, dfs(i, j));
      }
    }
    return ans;
  }

  private int dfs(int i, int j) {
    if (f[i][j] != 0) {
      return f[i][j];
    }
    int[] dirs = {-1, 0, 1, 0, -1};
    for (int k = 0; k < 4; ++k) {
      int x = i + dirs[k];
      int y = j + dirs[k + 1];
      if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
        f[i][j] = Math.max(f[i][j], dfs(x, y));
      }
    }
    return ++f[i][j];
  }
}

class Solution {
public:
  int longestIncreasingPath(vector<vector<int>>& matrix) {
    int m = matrix.size(), n = matrix[0].size();
    int f[m][n];
    memset(f, 0, sizeof(f));
    int ans = 0;
    int dirs[5] = {-1, 0, 1, 0, -1};

    function<int(int, int)> dfs = [&](int i, int j) -> int {
      if (f[i][j]) {
        return f[i][j];
      }
      for (int k = 0; k < 4; ++k) {
        int x = i + dirs[k], y = j + dirs[k + 1];
        if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
          f[i][j] = max(f[i][j], dfs(x, y));
        }
      }
      return ++f[i][j];
    };

    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        ans = max(ans, dfs(i, j));
      }
    }
    return ans;
  }
};

func longestIncreasingPath(matrix [][]int) (ans int) {
  m, n := len(matrix), len(matrix[0])
  f := make([][]int, m)
  for i := range f {
    f[i] = make([]int, n)
  }
  dirs := [5]int{-1, 0, 1, 0, -1}
  var dfs func(i, j int) int
  dfs = func(i, j int) int {
    if f[i][j] != 0 {
      return f[i][j]
    }
    for k := 0; k < 4; k++ {
      x, y := i+dirs[k], j+dirs[k+1]
      if 0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j] {
        f[i][j] = max(f[i][j], dfs(x, y))
      }
    }
    f[i][j]++
    return f[i][j]
  }
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      ans = max(ans, dfs(i, j))
    }
  }
  return
}

function longestIncreasingPath(matrix: number[][]): number {
  const m = matrix.length;
  const n = matrix[0].length;
  const f: number[][] = Array(m)
    .fill(0)
    .map(() => Array(n).fill(0));
  const dirs = [-1, 0, 1, 0, -1];
  const dfs = (i: number, j: number): number => {
    if (f[i][j] > 0) {
      return f[i][j];
    }
    for (let k = 0; k < 4; ++k) {
      const x = i + dirs[k];
      const y = j + dirs[k + 1];
      if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[i][j]) {
        f[i][j] = Math.max(f[i][j], dfs(x, y));
      }
    }
    return ++f[i][j];
  };
  let ans = 0;
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      ans = Math.max(ans, dfs(i, j));
    }
  }
  return ans;
}