392. Is Subsequence #

Problem #

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

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

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).

Example 1:

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

Example 2:

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

Follow up: If there are lots of incoming S, say S1, S2, … , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits: Special thanks to  @pbrother for adding this problem and creating all test cases.

Problem Summary #

Given strings s and t, determine whether s is a subsequence of t. You may assume that s and t contain only lowercase English letters. String t may be very long (length ~= 500,000), while s is a short string (length <=100). A subsequence of a string is a new string formed by deleting some (or none) of the characters from the original string without changing the relative positions of the remaining characters. (For example, “ace” is a subsequence of “abcde”, while “aec” is not.)

Solution Approach #

• Given 2 strings s and t, determine whether s is a subsequence of t. Note that s also needs to maintain the order of its letters in t.
• This is a greedy algorithm problem. Just solve it directly.

Code #

package leetcode

// Solution 1 O(n^2)
func isSubsequence(s string, t string) bool {
	index := 0
	for i := 0; i < len(s); i++ {
		flag := false
		for ; index < len(t); index++ {
			if s[i] == t[index] {
				flag = true
				break
			}
		}
		if flag == true {
			index++
			continue
		} else {
			return false
		}
	}
	return true
}

// Solution 2 O(n)
func isSubsequence1(s string, t string) bool {
	for len(s) > 0 && len(t) > 0 {
		if s[0] == t[0] {
			s = s[1:]
		}
		t = t[1:]
	}
	return len(s) == 0
}