Description
You are given two strings s and t consisting of only lowercase English letters.
Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.
A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.
Example 1:
Input: s = "coaching", t = "coding"
Output: 4
Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
Now, t is a subsequence of s ("coachingding").
It can be shown that appending any 3 characters to the end of s will never make t a subsequence.
Example 2:
Input: s = "abcde", t = "a"
Output: 0
Explanation: t is already a subsequence of s ("abcde").
Example 3:
Input: s = "z", t = "abcde"
Output: 5
Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
Now, t is a subsequence of s ("zabcde").
It can be shown that appending any 4 characters to the end of s will never make t a subsequence.
Constraints:
1 <= s.length, t.length <= 105sandtconsist only of lowercase English letters.
Solutions
Solution 1: Two Pointers
We define two pointers i and j, pointing to the first characters of strings s and t respectively. We iterate through string s, if s[i] = t[j], then we move j one step forward. Finally, we return n - j, where n is the length of string t.
The time complexity is O(m + n), where m and n are the lengths of strings s and t respectively. The space complexity is O(1).
class Solution: def appendCharacters(self, s: str, t: str) -> int: n, j = len(t), 0 for c in s: if j < n and c == t[j]: j += 1 return n - j(code-box)
