What is the time complexity of Is Subsequence?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Is Subsequence?

This problem uses Two Pointers, String, Dynamic Programming. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Is Subsequence?

Insight 1: The order of characters in `s` must be preserved in `t`. We can use two pointers to iterate through `s` and `t` simultaneously. Insight 2: We only need to increment the pointer for `s` when we find a matching character in `t`. The pointer for `t` always increments to scan all of its characters.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Is Subsequence - LeetCode 392 Solution

Q: How to solve Is Subsequence?

The provided solution uses a two-pointer approach. One pointer `i` tracks the index in string `s` and the other pointer `j` tracks the index in string `t`. The algorithm iterates through both strings. If `s[i]` matches `t[j]`, we increment `i` to check for the next character in `s`. Regardless of a match, we always increment `j` to move through `t`. The algorithm returns `true` if `i` reaches the end of `s`, indicating that all characters in `s` were found in `t` in the correct order, otherwise, it returns `false`.

Q: Which companies ask LeetCode 392. Is Subsequence in interviews?

This problem is frequently asked at Accenture, Electronic Arts, Nvidia, Pinterest, Qualcomm and 6 more companies.

Is Subsequence - Complete Solution Guide

Is Subsequence is LeetCode problem 392, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two strings s and t , return true if s is a subsequence of t , or false otherwise . A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of " a b c d e " while "aec" is not). Example 1: Input: s = "abc", t = "ahbgdc" Output: true Example 2: Input: s = "axc", t = "ahbgdc" Output: false Constraints: 0 <= s.length <=

Detailed Explanation

The problem asks us to determine if a string `s` is a subsequence of another string `t`. A subsequence is formed by deleting zero or more characters from the original string without changing the order of the remaining characters. For instance, "ace" is a subsequence of "abcde", but "aec" is not because the characters are not in the correct order.

Solution Approach

The provided solution uses a two-pointer approach. One pointer `i` tracks the index in string `s` and the other pointer `j` tracks the index in string `t`. The algorithm iterates through both strings. If `s[i]` matches `t[j]`, we increment `i` to check for the next character in `s`. Regardless of a match, we always increment `j` to move through `t`. The algorithm returns `true` if `i` reaches the end of `s`, indicating that all characters in `s` were found in `t` in the correct order, otherwise, it returns `false`.

Step-by-Step Algorithm

Step 1: Initialize two pointers, `i` to 0 (index for string `s`) and `j` to 0 (index for string `t`).
Step 2: Iterate through both strings using a `while` loop, continuing as long as `i` is within the bounds of `s` and `j` is within the bounds of `t`.
Step 3: Inside the loop, compare `s[i]` and `t[j]`. If they are equal, it means we have found the next character of the subsequence `s` in `t`. Increment `i`.
Step 4: Regardless of whether the characters match, increment `j` to move to the next character in `t`.
Step 5: After the loop finishes (either `i` reaches the end of `s` or `j` reaches the end of `t`), check if `i` is equal to the length of `s`. If it is, it means we have found all characters of `s` in `t` in the correct order, so return `true`. Otherwise, return `false`.

Key Insights

Insight 1: The order of characters in `s` must be preserved in `t`. We can use two pointers to iterate through `s` and `t` simultaneously.
Insight 2: We only need to increment the pointer for `s` when we find a matching character in `t`. The pointer for `t` always increments to scan all of its characters.
Insight 3: If we successfully iterate through all characters of `s`, then `s` is a subsequence of `t`. If we reach the end of `t` before finishing `s`, then `s` is not a subsequence of `t`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Two Pointers, String, Dynamic Programming.

Companies

Asked at: Accenture, Electronic Arts, Nvidia, Pinterest, Qualcomm, Salesforce, Tekion, Tesla, Tinkoff, Wix, Yandex.

Is Subsequence | LeetCode 392

Easy

Two Pointers String Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean isSubsequence(String s, String t) {
        int i = 0, j = 0;
        while (i < s.length() && j < t.length()) {
            if (s.charAt(i) == t.charAt(j)) {
                i++;
            }
            j++;
        }
        return i == s.length();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 392: Is Subsequence - Detailed explanation and solution approaches.

Topics

Two Pointers String Dynamic Programming

Companies

AccentureElectronic ArtsNvidiaPinterestQualcommSalesforceTekionTesla

Stats

Difficulty:Easy

Languages:4

Companies:11

Problem #:392

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

5 steps

1
Step 1: Initialize two pointers, `i` to 0 (index for string `s`) and `j` to 0 (index for string `t`).
2
Step 2: Iterate through both strings using a `while` loop, continuing as long as `i` is within the bounds of `s` and `j` is within the bounds of `t`.
3
Step 3: Inside the loop, compare `s[i]` and `t[j]`. If they are equal, it means we have found the next character of the subsequence `s` in `t`. Increment `i`.
4
Step 4: Regardless of whether the characters match, increment `j` to move to the next character in `t`.
5
Step 5: After the loop finishes (either `i` reaches the end of `s` or `j` reaches the end of `t`), check if `i` is equal to the length of `s`. If it is, it means we have found all characters of `s` in `t` in the correct order, so return `true`. Otherwise, return `false`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the `while` loop iterates at most `n` times (where n is the length of `t`). The space complexity is O(1) because only a fixed number of variables are used regardless of the input string sizes.

📖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

Dynamic Programming

A dynamic programming solution can be constructed to solve this problem. A 2D table is created where `dp[i][j]` indicates whether `s[0...i]` is a subsequence of `t[0...j]`. This approach fills the table row by row.

Time: O(m*n), where m is the length of s and n is the length of t.Space: O(m*n)

Trade-offs: The two-pointer approach is more efficient in terms of both time and space for this specific problem. Dynamic programming can be useful for similar but more complex problems involving subsequences, particularly where some form of optimization needs to be calculated rather than simple verification. This DP approach is only useful when you need to keep track of the subsequences. For only validation, two pointer is optimal.

✨

Key Insights

1Insight 1: The order of characters in `s` must be preserved in `t`. We can use two pointers to iterate through `s` and `t` simultaneously.
2Insight 2: We only need to increment the pointer for `s` when we find a matching character in `t`. The pointer for `t` always increments to scan all of its characters.
3Insight 3: If we successfully iterate through all characters of `s`, then `s` is a subsequence of `t`. If we reach the end of `t` before finishing `s`, then `s` is not a subsequence of `t`.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

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

Edge Cases

Edge case 1: `s` is an empty string. In this case, the `while` loop will not execute and `i` will remain 0. The algorithm will return `i == len(s)`, which is `0 == 0`, so it will return `true`. An empty string is a subsequence of any string.
Edge case 2: `t` is an empty string and `s` is not an empty string. In this case, the `while` loop will not execute because `j < len(t)` will be false immediately. The algorithm will return `i == len(s)`, which will be `0 == len(s)`. Since `s` is not an empty string, `len(s)` is greater than 0, so the algorithm will return `false`.
Edge case 3: `s` and `t` are the same string. The algorithm correctly identifies `s` as a subsequence of `t` and returns `true`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that the characters of `s` must appear consecutively in `t`. Remember that characters can be skipped in `t`.

✗Common mistake 2: Forgetting to increment the pointer for `t` even when there is no match. The `j` pointer always moves forward to check all the characters of `t`.

✅

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 editors and search functionalities. Determining if a user's search query is a subsequence of the text in a document can help in providing fuzzy or partial match results.

🏦Real-world scenario 2: Bioinformatics. Identifying if a DNA sequence is a subsequence of a longer DNA strand can be crucial in identifying gene segments or patterns.

💡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

The two-pointer technique is powerful for sorted arrays. The key insight is recognizing when elements can be processed from both ends simultaneously.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

The most common mistake is not handling the case when pointers meet. Always consider the termination condition.

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: July 3, 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

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Is Subsequence"?

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

🤖 Need help with this problem?