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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 <= 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 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 \(O(1)\).

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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

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

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

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func appendCharacters(s string, t string) int {
    n, j := len(t), 0
    for _, c := range s {
        if j < n && byte(c) == t[j] {
            j++
        }
    }
    return n - j
}

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function appendCharacters(s: string, t: string): number {
    let j = 0;
    for (const c of s) {
        if (c === t[j]) {
            ++j;
        }
    }
    return t.length - j;
}

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