Leetcode #291: Word Pattern II
In this guide, we solve Leetcode #291 Word Pattern II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a pattern and a string s, return true if s matches the pattern. A string s matches a pattern if there is some bijective mapping of single characters to non-empty strings such that if each character in pattern is replaced by the string it maps to, then the resulting string is s.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Hash Table, String, Backtracking
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: pattern = "abab", s = "redblueredblue"
Output: true
Explanation: One possible mapping is as follows:
'a' -> "red"
'b' -> "blue"
Python Solution
class Solution:
def wordPatternMatch(self, pattern: str, s: str) -> bool:
def dfs(i, j):
if i == m and j == n:
return True
if i == m or j == n or n - j < m - i:
return False
for k in range(j, n):
t = s[j : k + 1]
if d.get(pattern[i]) == t:
if dfs(i + 1, k + 1):
return True
if pattern[i] not in d and t not in vis:
d[pattern[i]] = t
vis.add(t)
if dfs(i + 1, k + 1):
return True
d.pop(pattern[i])
vis.remove(t)
return False
m, n = len(pattern), len(s)
d = {}
vis = set()
return dfs(0, 0)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.