Leetcode #792: Number of Matching Subsequences
In this guide, we solve Leetcode #792 Number of Matching Subsequences 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 an array of strings words, return the number of words[i] that is a subsequence of s. A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Trie, Array, Hash Table, String, Binary Search, Dynamic Programming, Sorting
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 = "abcde", words = ["a","bb","acd","ace"]
Output: 3
Explanation: There are three strings in words that are a subsequence of s: "a", "acd", "ace".
Python Solution
class Solution:
def numMatchingSubseq(self, s: str, words: List[str]) -> int:
d = defaultdict(deque)
for w in words:
d[w[0]].append(w)
ans = 0
for c in s:
for _ in range(len(d[c])):
t = d[c].popleft()
if len(t) == 1:
ans += 1
else:
d[t[1]].append(t[1:])
return 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.