# 2486. Append Characters to String to Make Subsequence

## 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 <= 105
• s and t consist 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 traverse string $t$, when $s[i] \neq t[j]$, we move pointer $i$ forward until $s[i] = t[j]$ or $i$ reaches the end of string $s$. If $i$ reaches the end of string $s$, it means that the character $t[j]$ in $t$ cannot find the corresponding character in $s$, so we return the remaining number of characters in $t$. Otherwise, we move both pointers $i$ and $j$ forward and continue to traverse string $t$.

The time complexity is $O(m + n)$, and the space complexity is $O(1)$. Where $m$ and $n$ are the lengths of strings $s$ and $t$ respectively.

• class Solution {
public int appendCharacters(String s, String t) {
int m = s.length(), n = t.length();
for (int i = 0, j = 0; j < n; ++j) {
while (i < m && s.charAt(i) != t.charAt(j)) {
++i;
}
if (i++ == m) {
return n - j;
}
}
return 0;
}
}

• class Solution {
public:
int appendCharacters(string s, string t) {
int m = s.size(), n = t.size();
for (int i = 0, j = 0; j < n; ++j) {
while (i < m && s[i] != t[j]) {
++i;
}
if (i++ == m) {
return n - j;
}
}
return 0;
}
};

• class Solution:
def appendCharacters(self, s: str, t: str) -> int:
i, m = 0, len(s)
for j, c in enumerate(t):
while i < m and s[i] != c:
i += 1
if i == m:
return len(t) - j
i += 1
return 0


• func appendCharacters(s string, t string) int {
m, n := len(s), len(t)
for i, j := 0, 0; j < n; i, j = i+1, j+1 {
for i < m && s[i] != t[j] {
i++
}
if i == m {
return n - j
}
}
return 0
}

• function appendCharacters(s: string, t: string): number {
const [m, n] = [s.length, t.length];
for (let i = 0, j = 0; j < n; ++j) {
while (i < m && s[i] !== t[j]) {
++i;
}
if (i === m) {
return n - j;
}
++i;
}
return 0;
}
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