Published: May 24, 2024
Problem Description
You are given two strings
s
andt
consisting of only lowercase English letters.Return the minimum number of characters that need to be appended to the end of
s
so thatt
becomes a subsequence ofs
.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
andt
consist only of lowercase English letters.https://leetcode.com/problems/append-characters-to-string-to-make-subsequence/
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;
}
};
Complexities
- Time:
O(n)
- Space:
O(1)