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Problem solution · Python

Sequential Grid Path Cover

Sequential Grid Path Cover: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Depth-first search
Source
Kamyu LeetCode Solutions
Length
38 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Sequential Grid Path Cover, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 38 lines of Python from the credited upstream file sequential-grid-path-cover.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSequential Grid Path Cover · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(m * n * 3^(m * n))# Space: O(m * n) # backtrackingclass Solution(object):    def findPath(self, grid, k):        """        :type grid: List[List[int]]        :type k: int        :rtype: List[List[int]]        """        DIRECTIONS = ((1, 0), (0, 1), (-1, 0), (0, -1))        def backtracking(i, j, curr):            v = grid[i][j]            if v and v != curr:                return False            grid[i][j] = -1            result.append([i, j])            if len(result) == len(grid)*len(grid[0]):                return True            new_curr = curr+1 if v == curr else curr            for di, dj in DIRECTIONS:                ni, nj = i+di, j+dj                if not (0 <= ni < len(grid) and 0 <= nj < len(grid[0]) and grid[ni][nj] != -1):                    continue                if backtracking(ni, nj, new_curr):                    return True            result.pop()            grid[i][j] = v            return False         result = []        for i in xrange(len(grid)):            for j in xrange(len(grid[0])):                if backtracking(i, j, 1):                    return result        return result

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