Let's solve LeetCode - Is Subsequence

Problem 392 - Is Subsequence

Given a string s and a string t, check if s is a subsequence of t.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Examples

Input: s = "abc", t = "ahbgdc"
Output: true

Input: s = "axc", t = "ahbgdc"
Output: false

Conceptual Overview

Since we want to check if s is a subsequence of t we will want to check every character of s against t and if a character at s matches a character in t (in order) then we can move onto the next character in s and do the check all over again.

Looking at the example above let's run through a couple of iterations. In ECMAScript 5 we can treat the string as an array-like object, where individual characters correspond to a numerical index.

1) At s[0] = a; t[0] = a; does s[0] === t[0]? Yes, move to the next character in s and t
2) At s[1] = b; t[1] = h; does s[1] === t[0]? No, move to the next character in t
3) At s[1] = b; t[2] = b; does s[1] === t[2]? Yes, move to the next character in s and t
...
6) At s[2] = c; t[5] = c; does s[3] === t[5]? Yes, and since we went through the length of s we found s to be a subsequence of t

Code

While-loop variation

/**
 * @param {string} s
 * @param {string} t
 * @return {boolean}
 */
const isSubsequence = (s, t) => {
    if (s.length === 0) return true

    let sPointer = 0
    let tPointer = 0

    while (sPointer < s.length && tPointer < t.length) {
        if(s[sPointer] === t[tPointer]) sPointer++

        tPointer++
    }

    return sPointer === s.length

};

For-loop variation

const isSubsequence = (s, t) => {
    if (s.length === 0) return true

    let sPointer = 0

    for (let i = 0; i < t.length; i++) {
        if (s[sPointer] === t[i]) sPointer++
    }
    return sPointer === s.length
}

In both code solutions, we need to keep track of our position in each string so we use pointers to help with that.

Time & Space Complexity

Time: O(n) - where n is the length of the array
Space: O(1)
Both variations have the same time and space complexity