392. Is Subsequence

 

Building an Efficient Subsequence Checker in JavaScript

When working with strings, one common task is to determine if one string is a subsequence of another. A subsequence maintains the relative order of characters but does not require them to be contiguous. In this blog post, we'll walk through an efficient implementation of a subsequence checker in JavaScript.

Problem Statement

Given two strings s and t, the goal is to determine if s is a subsequence of t. A subsequence of a string is formed by deleting some (or no) characters from the string without changing the order of the remaining characters.

Example

  • Input: s = "abc", t = "ahbgdc"

  • Output: true

  • Input: s = "axc", t = "ahbgdc"

  • Output: false

Intuition

To determine if s is a subsequence of t, we need to verify if the characters in s appear in t in the same order. This can be achieved using a simple iteration approach combined with a stack-like behavior.

Approach

  1. Convert s to an Array: Convert the string s into an array of characters. This allows us to use array operations like pop.

  2. Iterate Through t from End to Start: Use a loop to iterate through the string t from the last character to the first.

  3. Check for Matching Characters: For each character in t, if it matches the last character of the s array, remove that character from s using s.pop().

  4. Return Result: After iterating through t, if the s array is empty, it means all characters of s have been found in t in the correct order, and we return true. Otherwise, return false.

Code Implementation

Here's how you can implement the subsequence checker in JavaScript:

javascript
var isSubsequence = function(s, t) {
    s = Array.from(s);
    for(let i = t.length - 1; i >= 0; i--) {
        if(t[i] === s[s.length - 1]) {
            s.pop();
        }
    }
    return s.length ? false : true;
};

Complexity Analysis

  • Time Complexity: O(n), where n is the length of the string t. This is because we iterate through t once.

  • Space Complexity: O(m), where m is the length of the string s. This is because we convert s to an array, which takes up additional space.

Conclusion

This solution provides an efficient way to determine if a string s is a subsequence of another string t. By using a simple iteration and array manipulation, we can ensure that the characters of s appear in the correct order within t. This approach is straightforward and handles the problem effectively.

If you're looking for more detailed explanations and additional solutions, check out the discussion on LeetCode: https://leetcode.com/problems/is-subsequence/solutions/6143778/two-pointer-easy-solution-by-bikash_kuma-f66r/

Happy coding! šŸ˜Š

Comments

Post a Comment

Popular posts from this blog

3Sum Closest: O(n²) Two-Pointer Magic šŸš€ leetcode: 16

Three Sum Closest - LeetCode Solution In this blog post, we'll dive into the Three Sum Closest problem, a popular coding challenge on LeetCode. We'll explore the intuition behind the solution, discuss the approach, analyze the complexity, and provide a detailed explanation of the code. Additionally, I'll share my LeetCode solution link for further reference. Problem Statement Given an integer array nums of length n and an integer target , find three integers in nums such that the sum is closest to target . Return the sum of the three integers. You may assume that each input would have exactly one solution. Intuition The problem requires finding three numbers in the array whose sum is as close as possible to the target. To efficiently find such a triplet, we can use a combination of sorting and the two-pointer technique . Sorting : First, sort the array. This allows us to use the two-po...
Read more

Kadane's Algorithm: Maximum Subarray Problem - LeetCode(53) Solution

Maximum Subarray Problem - LeetCode Solution Maximum Subarray Problem - LeetCode Solution Problem Statement Given an integer array nums , find the contiguous subarray (containing at least one number) that has the largest sum and return its sum. Intuition The problem requires finding the contiguous subarray with the maximum sum. A brute force approach would involve checking all subarrays, but this would be inefficient. Instead, we use Kadane's Algorithm , which efficiently finds the maximum sum subarray in O(n) time. Approach Initialize two variables: total (running sum) and maxi (maximum sum encountered so far). Iterate through the array: Add the current element to total . Update maxi with the maximum of maxi and total . If total becomes negative, reset it to zero. Return maxi as the maximum sub...
Read more

LeetCode Problem: Implement pow(x, n) leetcode:- 50

Understanding the Fast Exponentiation Algorithm: Exponentiation by Squaring In this blog post, we will explore the implementation of the myPow function, a common problem found on LeetCode. The goal of this problem is to compute \(x^n\) efficiently, even when \(n\) is a very large number. Instead of using a naive approach, we employ a method called Exponentiation by Squaring to significantly reduce the number of multiplications required. Problem Overview Given a floating-point number x and an integer n , our task is to implement a function that computes \(x^n\). A straightforward solution might involve multiplying x by itself n times, but this approach is too slow when n is large. Intuition Behind Exponentiation by Squaring The main idea is to reduce the problem size by halving the exponent at each recursive step. Consider the following: If n is even, then: \(x^n = x^{...
Read more