Leetcode #1178: Number of Valid Words for Each Puzzle
In this guide, we solve Leetcode #1178 Number of Valid Words for Each Puzzle 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
With respect to a given puzzle string, a word is valid if both the following conditions are satisfied: word contains the first letter of puzzle. For each letter in word, that letter is in puzzle.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Trie, Array, Hash Table, String
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: words = ["aaaa","asas","able","ability","actt","actor","access"], puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
Output: [1,1,3,2,4,0]
Explanation:
1 valid word for "aboveyz" : "aaaa"
1 valid word for "abrodyz" : "aaaa"
3 valid words for "abslute" : "aaaa", "asas", "able"
2 valid words for "absoryz" : "aaaa", "asas"
4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
There are no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.
Python Solution
class Solution:
def findNumOfValidWords(self, words: List[str], puzzles: List[str]) -> List[int]:
cnt = Counter()
for w in words:
mask = 0
for c in w:
mask |= 1 << (ord(c) - ord("a"))
cnt[mask] += 1
ans = []
for p in puzzles:
mask = 0
for c in p:
mask |= 1 << (ord(c) - ord("a"))
x, i, j = 0, ord(p[0]) - ord("a"), mask
while j:
if j >> i & 1:
x += cnt[j]
j = (j - 1) & mask
ans.append(x)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.