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Problem solution · Python

Word Pattern II

Word Pattern II: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
42 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 Word Pattern II, 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

  • 42 lines of Python from the credited upstream file 291.py.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • No explicit loop blocks detected, together with recursive traversal.

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 walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeWord Pattern II · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def wordPatternMatch(self, pattern: str, s: str) -> bool:    def isMatch(            i: int, j: int, charToString: dict[str, str],            seen: set[str]) -> bool:      if i == len(pattern) and j == len(s):        return True      if i == len(pattern) or j == len(s):        return False       c = pattern[i]       if c in charToString:        t = charToString[c]        # See if we can match t with s[j..n).        if t not in s[j:]:          return False         # If there's a match, continue to match the rest.        return isMatch(i + 1, j + len(t), charToString, seen)       for k in range(j, len(s)):        t = s[j:k + 1]         # This string is mapped by another character.        if t in seen:          continue         charToString[c] = t        seen.add(t)         if isMatch(i + 1, k + 1, charToString, seen):          return True         # Backtrack.        del charToString[c]        seen.remove(t)       return False     return isMatch(0, 0, {}, set())

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