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Nilesh Raut
Nilesh Raut

Posted on • Updated on • Originally published at nileshblog.tech

Cracking the LeetCode:392 Is Subsequence?

What is LeetCode 392: Is Subsequence?
Before we dive into the intricacies of LeetCode 392, let's grasp the concept. This coding challenge, categorized as "easy," tests our ability to determine if one string is a subsequence of another. Think of it as a detective's quest to find clues in a larger mystery.
Unpacking the Challenge
Imagine you have two strings: s and t. Your task is to check whether s is a subsequence of t. In other words, can we derive string s by deleting some (or none) characters from string t while maintaining the order of characters in s? If we can, the answer is "yes"; otherwise, it's "no."
To illustrate this, let's look at an example:
Example:

  • s: "abc"
  • t: "ahbgdc" In this case, the answer is "yes" because we can delete characters from t to form s. We can remove 'h' and 'g,' leaving us with "abc." The Approach: Greedy is the Key To tackle this problem efficiently, we'll adopt a greedy approach. Think of it as picking the ripest fruit from a tree. Here's the plan:
  • Initialize two pointers, one for each string: i for s and j for t.
  • Traverse both strings using these pointers.
  • Compare characters at the current positions of i and j.
  • If they match, move both pointers forward.
  • If they don't match, only move the j pointer.
  • Keep doing this until you reach the end of either string. If you successfully traverse the entire s string, it means s is a subsequence of t. If the j pointer reaches the end of t before i finishes with s, then it's not a subsequence. Time Complexity Matters Our approach is efficient with a time complexity of O(m + n), where m is the length of s and n is the length of t. We traverse both strings only once, making this a linear time solution. Coding It Out Let's put theory into practice with some Python code: pythonCopy codedef isSubsequence(s, t): 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)

Conclusion: Cracked the Code
LeetCode 392. Is Subsequence ," is a problem that can be efficiently solved using a simple greedy approach. By comparing characters of both strings, we can determine whether one is a subsequence of the other. Remember, it's not about brute force; it's about being smart and efficient, just like a detective solving a case.
So, the next time you encounter this problem, don't be intimidated. You've got the code-cracking skills to answer, "Is Subsequence?" with confidence. Happy coding!

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