Is Subsequence Leetcode Problem 392 [Python Solution]

In this blog post, we will explore the LeetCode problem Is Subsequence This is an easy problem that falls under the category of Arrays & Hashing and is frequently encountered in coding interviews, especially in the context of dynamic programming.

We'll provide a step-by-step explanation of the problem, discuss the constraints, and walk you through both the brute-force and efficient approaches to solving it.

Problem Overview

The problem statement is as follows: Given two strings, s and t, we need to determine whether s is a subsequence of t.

A subsequence of a string is a new string that can be formed from the original string by deleting some (or none) of the characters without changing the relative positions of the remaining characters.

In other words, a subsequence is a string formed by selecting characters from the original string without changing their order.

For example:
– If s = "abc" and t = "ahbgdc", the function should return True because we can form s by selecting characters from t (e.g., "a" from "ahbgdc," "b" from "b," and "c" from "c").
– If s = "axc" and t = "ahbgdc", the function should return False because "axc" cannot be formed by selecting characters from "ahbgdc" in the required order.

Constraints

Before we dive into the solution, it's essential to understand the constraints of the problem.

These constraints are crucial in designing an efficient algorithm.

• 0 <= s.length <= 100
• 0 <= t.length <= 10,000
• Both s and t consist only of lowercase English letters.

Now that we have a clear understanding of the problem and its constraints, let's discuss the approaches to solving it.

Brute-force Approach

The brute-force approach is straightforward and involves comparing each character of s with characters in t while keeping track of two pointers.

Here's a step-by-step explanation:

1. Initialize two pointers, i and j, at the beginning of strings s and t, respectively.

2. While both pointers are within bounds (i.e., i < len(s) and j < len(t)), compare the characters s[i] and `t[j].

3. If the characters are equal, increment both pointers (i and j).

This signifies that we found a matching character in both strings.

1. If the characters are not equal, increment only the pointer for string t (j).

This indicates that we didn't find a match for the character in s.

1. Continue this process until i reaches the end of s or until one of the pointers goes out of bounds.

2. If i reaches the end of s, it means we found all characters in s in the required order within t, and we return True.

3. If the loop ends before finding all characters in s, we return False.

The code implementation for the brute-force approach is as follows:

def isSubsequence(s: str, t: str) -> bool:
    i, j = 0, 0
    while i < len(s) and j < len(t):
        if s[i] == t[j]:
            i += 1
        j += 1
    return i == len(s)

This code provides a clear and straightforward way to solve the problem, and it operates in linear time complexity, making it quite efficient.

Efficient Approach

The brute-force approach is efficient, but we can optimize it further.

We don't need to compare each character of s with t in a linear manner.

Instead, we can take advantage of Python's built-in functions.

The more efficient approach uses Python's str.find() method.

This method searches for the first occurrence of a substring in a string.

We iteratively search for each character of s in t while maintaining the search starting point.

If the character is found, we update the starting point for the next search to the position after the found character.

If any character is not found, we return False.

If we successfully find all characters in s, we return True.

Here's the efficient code implementation:

def isSubsequence(s: str, t: str) -> bool:
    start = 0
    for char in s:
        start = t.find(char, start)
        if start == -1:
            return False
        start += 1
    return True

This code is more concise and efficient, as it takes advantage of Python's built-in functions to simplify the process of finding characters in t.

It also operates in linear time complexity, making it a faster solution.

Time and Space Complexity

Time Complexity

• The brute-force approach and the efficient approach both have a time complexity of O(N), where N is the sum of the lengths of s and t.

This is because, in the worst case, we may have to iterate through both strings once.

Space Complexity

• Both approaches have a space complexity of O(1) as they use a constant amount of extra space, regardless of the input size.

Reasoning Behind Our Approach

The reasoning behind both approaches is straightforward.

We need to compare two strings and determine if one is a subsequence of the other.

The brute-force approach uses two pointers to perform this comparison, while the efficient approach leverages Python's built-in str.find() method to make the process more efficient.

In the brute-force approach, we compare characters one by one, and if we find a matching character, we move both pointers to the next position.

If we don't find a match, we only move the pointer in t.

If we successfully reach the end of s, it means s is a subsequence of t.

In the efficient approach, we iteratively use str.find() to search for each character of s in t.

If we find a character, we update the starting point for the next search.

If we don't find a character, we return False.

This approach simplifies the code and operates in the same linear time complexity as the brute-force approach.

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Conclusion

In this blog post, we explored the LeetCode problem Is Subsequence and provided both a brute-force and an efficient solution to determine if one string is a subsequence of another.

We discussed the problem's constraints, explained the reasoning behind our approaches, and analyzed the time and space complexities.

The brute-force approach involves comparing characters one by one, while the efficient approach leverages Python's str.find() method to optimize the process.

Both solutions operate in linear time complexity, making them efficient for solving the problem.

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If you have any more questions or if you'd like to practice this problem, you can find it on LeetCode here.

Happy coding!