Append Characters to String to Make Subsequence

Published: May 24, 2024

Problem 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.

Constraints:

• 1 <= s.length, t.length <= 10**5
• s and t consist only of lowercase English letters.

Examples

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.

How to Solve

The question asks the length of t’s postfix. Use two pointers to identify a prefix. One goes through s, another goes through t sequentially. Increment s’s index one by one. Only when two characters in s and t are the same, increment t’s index. When the loop is over, t’s length minus ‘t`’s index is the answer.

Solution

class AppendCharsToMakeSubsequence {
public:
    int appendCharacters(string s, string t) {
        std::ios_base::sync_with_stdio(false);
        std::cin.tie(NULL);
        
        int m = s.size(), n = t.size(), j = 0;
        for (int i = 0; i < m && j < n; ++i) {
            if (s[i] == t[j]) j++;
        }
        return n - j;
    }
};

# @param {String} s
# @param {String} t
# @return {Integer}
def append_characters(s, t)
    m, n, i, j = s.length, t.length, 0, 0
    while i < m && j < n
      if s[i] == t[j]
        j += 1
      end
      i += 1
    end
    n - j
end

Complexities

• Time: O(n)
• Space: O(1)