Leetcode #140: Word Break II
In this guide, we solve Leetcode #140 Word Break 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 string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Trie, Memoization, Array, Hash Table, String, Dynamic Programming, 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: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
Output: ["cats and dog","cat sand dog"]
Python Solution
class Trie:
def __init__(self):
self.children = [None] * 26
self.is_end = False
def insert(self, word):
node = self
for c in word:
idx = ord(c) - ord('a')
if node.children[idx] is None:
node.children[idx] = Trie()
node = node.children[idx]
node.is_end = True
def search(self, word):
node = self
for c in word:
idx = ord(c) - ord('a')
if node.children[idx] is None:
return False
node = node.children[idx]
return node.is_end
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
def dfs(s):
if not s:
return [[]]
res = []
for i in range(1, len(s) + 1):
if trie.search(s[:i]):
for v in dfs(s[i:]):
res.append([s[:i]] + v)
return res
trie = Trie()
for w in wordDict:
trie.insert(w)
ans = dfs(s)
return [' '.join(v) for v in ans]
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.