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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 <= 10^{5}
s
andt
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; }