What is the time complexity of Number of Matching Subsequences?

The optimal solution has a time complexity of O(S+W*L).

What is the space complexity of Number of Matching Subsequences?

The space complexity is O(W*L).

What data structures are used to solve Number of Matching Subsequences?

This problem uses Array, Hash Table, String, Binary Search, Dynamic Programming, Trie, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Matching Subsequences?

Insight 1: Iterating through the main string `s` and efficiently checking if each word in `words` is a subsequence is crucial. Insight 2: Using iterators to track the current position in each word can help efficiently verify the subsequence property.

Which companies ask LeetCode 792. Number of Matching Subsequences in interviews?

This problem is frequently asked at Salesforce, Visa.

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Number of Matching Subsequences - LeetCode 792 Solution

Number of Matching Subsequences - Complete Solution Guide

Number of Matching Subsequences is LeetCode problem 792, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. For example, "ace" is a subsequence of "abcde" . Example 1: 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",

Detailed Explanation

The problem asks us to find the number of strings in a given array of strings (`words`) that are subsequences of another given string (`s`). A subsequence is formed by deleting zero or more characters from the original string without changing the relative order of the remaining characters. For instance, "ace" is a subsequence of "abcde". The inputs are the string `s` and the array of strings `words`. The output is an integer representing the count of strings in `words` that are subsequences of `s`. The constraints specify the lengths of the strings and arrays, limiting the size of the problem.

Solution Approach

The solution uses a hash table (specifically, a `defaultdict` in Python or a `HashMap` in Java/C++) to group the words based on their current expected character. Initially, words are grouped based on their first character. The algorithm iterates through the characters of the string `s`. For each character in `s`, it checks if there are any words waiting for that character in the hash table. If found, it processes those words by advancing their iterators to the next character they need to match. If a word's iterator reaches the end, it means the entire word is a subsequence of `s`, and the count is incremented. Words that are still subsequences get re-added to the hash table under the next expected character.

Step-by-Step Algorithm

Step 1: Create a hash table (or dictionary) called `waiting` to store lists of iterators, keyed by the character they're waiting for.
Step 2: Iterate through the `words` array. For each word, create an iterator and add it to the `waiting` list corresponding to its first character.
Step 3: Initialize a counter `count` to 0. This will keep track of the number of subsequences found.
Step 4: Iterate through the string `s`. For each character `char` in `s`:
Step 5: Get the list of iterators waiting for `char` from the `waiting` hash table. Remove this entry from the table.
Step 6: Iterate through the retrieved list of iterators. For each iterator `it`:
Step 7: Advance the iterator `it` to its next character. If `it` reaches the end of the word (i.e., no next character), increment `count` because the current word is a subsequence of `s`.
Step 8: If `it` has a next character, add it to the `waiting` list associated with that character.
Step 9: After iterating through the entire string `s`, return the `count`.

Key Insights

Insight 1: Iterating through the main string `s` and efficiently checking if each word in `words` is a subsequence is crucial.
Insight 2: Using iterators to track the current position in each word can help efficiently verify the subsequence property.
Insight 3: Employing a hash table (or dictionary) to group words by their starting characters allows for efficient processing during iteration of `s`.

Complexity Analysis

Time Complexity: O(S+W*L)

Space Complexity: O(W*L)

Topics

This problem involves: Array, Hash Table, String, Binary Search, Dynamic Programming, Trie, Sorting.

Companies

Asked at: Salesforce, Visa.

Number of Matching Subsequences | LeetCode 792

Medium

Array Hash Table String Binary Search Dynamic Programming Trie Sorting

Time: O(S+W*L)Space: O(W*L)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;

class Solution {
    public int numMatchingSubseq(String s, String[] words) {
        Map<Character, List<Iterator<Character>>> waiting = new HashMap<>();
        for (String w : words) {
            Iterator<Character> it = w.chars().mapToObj(c -> (char) c).iterator();
            // The problem constraints state 1 <= words[i].length, so it.next() is safe.
            if (it.hasNext()) {
                char firstChar = it.next();
                if (!waiting.containsKey(firstChar)) {
                    waiting.put(firstChar, new ArrayList<>());
                }
                waiting.get(firstChar).add(it);
            }

        }

        int count = 0;
        for (char charS : s.toCharArray()) {
            // Get all iterators waiting for the current character.
            // We use remove to remove the entry from the dictionary, as these
            // iterators will be advanced to wait for their next character.
            List<Iterator<Character>> advancing = waiting.remove(charS);
            if (advancing != null) {
                for (Iterator<Character> it : advancing) {
                    // Advance the iterator.
                    if (it.hasNext()) {
                        char nextChar = it.next();
                        if (nextChar != Character.MIN_VALUE) {
                            // If there's a next character, add the iterator to the
                            // waiting list for that character.
                            if (!waiting.containsKey(nextChar)) {
                                waiting.put(nextChar, new ArrayList<>());
                            }
                            waiting.get(nextChar).add(it);
                        }
                    } else {
                        // If the iterator is exhausted, the word is a subsequence.
                        count += 1;
                    }
                }
            }
        }

        return count;
    }
}

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Problem Info

Description

Problem 792: Number of Matching Subsequences - Detailed explanation and solution approaches.

Topics

Array Hash Table String Binary Search Dynamic Programming Trie Sorting

Companies

SalesforceVisa

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:792

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Create a hash table (or dictionary) called `waiting` to store lists of iterators, keyed by the character they're waiting for.
2
Step 2: Iterate through the `words` array. For each word, create an iterator and add it to the `waiting` list corresponding to its first character.
3
Step 3: Initialize a counter `count` to 0. This will keep track of the number of subsequences found.
4
Step 4: Iterate through the string `s`. For each character `char` in `s`:
5
Step 5: Get the list of iterators waiting for `char` from the `waiting` hash table. Remove this entry from the table.
6
Step 6: Iterate through the retrieved list of iterators. For each iterator `it`:
7
Step 7: Advance the iterator `it` to its next character. If `it` reaches the end of the word (i.e., no next character), increment `count` because the current word is a subsequence of `s`.
8
Step 8: If `it` has a next character, add it to the `waiting` list associated with that character.
9
Step 9: After iterating through the entire string `s`, return the `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(S+W*L)

💾

Space Complexity

O(W*L)

📝

Why?

The time complexity comes from iterating through `s` once (O(S)). For each character in `s`, we iterate through a subset of words (in the worst case, all words W). For each word, we advance the iterator a maximum of L times (the length of the word). Thus, the complexity can be expressed as O(S + W * L). The space complexity arises primarily from the `waiting` hash map. In the worst-case scenario, all words might be waiting for a particular character at some point, so the space required would be proportional to the total length of all words which would be O(W*L).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Two Pointers

For each word, use two pointers, one for `s` and one for the word. Advance the `s` pointer until a match is found. If the word pointer reaches the end, the word is a subsequence.

Time: O(S*W*L) in worst case where you traverse s for each element of W.Space: O(1)

Trade-offs: Less space complexity. Potentially slower than the main approach, especially when words share common prefixes. More intuitive than the iterator approach for beginners. Less efficient for large word arrays because it would re-iterate the string 's' multiple times.

✨

Key Insights

1Insight 1: Iterating through the main string `s` and efficiently checking if each word in `words` is a subsequence is crucial.
2Insight 2: Using iterators to track the current position in each word can help efficiently verify the subsequence property.
3Insight 3: Employing a hash table (or dictionary) to group words by their starting characters allows for efficient processing during iteration of `s`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `words` array is empty. The algorithm correctly returns 0 as no subsequences can be found.
Edge case 2: `s` is an empty string. The algorithm will iterate zero times, so the waiting array will never be cleared which results in no count increment.
Edge case 3: A word in `words` is an empty string. While the problem statement implies the word lengths are at least 1, the logic handles this case, because a `next` call on the iterator of an empty string will immediately evaluate as false, incrementing the count.
Edge case 4: words in `words` is very large and long, but the alphabet is small, causing many collisions in the hash table. It will still execute as intended, only there may be more loops due to the bucket sizes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where a word is fully consumed (iterator reaches the end). This will lead to an incorrect count.

✗Common mistake 2: Forgetting to remove the entry from the hash table when processing words waiting for a particular character. This will cause the same words to be processed multiple times.

✗Common mistake 3: Modifying the `words` array directly instead of using iterators. This makes the logic complex to manage and debug.

✗Common mistake 4: Improper handling of the case where a word in `words` is not a subsequence, leading to it being incorrectly counted.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text editing/searching: Finding if a user's input (words) is a subsequence of a document (s). For example, type ahead features or substring matches.

🏦Real-world scenario 2: Bioinformatics: Identifying if a DNA sequence (words) is a subsequence of a longer DNA sequence (s).

🎮Real-world scenario 3: Data streaming: Given a stream of data (s), determine if certain patterns (words) exist as subsequences.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Matching Subsequences"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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