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This approach uses two pointers to scan both strings, one pointer for each string. The idea is to attempt to match characters of t with those in s. A character from t is successfully matched if it is equal to the current character in s. We iterate through s, moving the pointer for each character matched, while the pointer for t only moves if a match is found. Once we've iterated through s, we determine how many characters are left unmatched in t, which is the number of characters needed to be appended to s.
Time Complexity: O(n + m) where n is the length of s and m is the length of t.
Space Complexity: O(1) as no additional space is used.
1public class MinCharsAppend {
2 public static int minCharsToAppend(String s, String t) {
3 int sLen = s.length();
4
The Java solution defines a function minCharsToAppend that iterates over the strings using two indices. The index for t only increments when a matching character is found in s. At the end of the iteration, it calculates how many characters of t are not matched and need to be appended.
This approach uses dynamic programming to precompute the next occurrence of each character for every position in the string s. By doing so, we quickly identify the existence and location of each character from t in s. We build a 2D array next where next[i][j] represents the smallest index >= i that holds the character j in s. This helps in efficiently determining the position of characters and the necessary additions to the end of the string.
Time Complexity: O(n + m + n * ALPHABET_SIZE) due to preprocessing each character for positions.
Space Complexity: O(n * ALPHABET_SIZE) used by the 2D array next where n is length of s.
1using System;
2
3public class MinChars
4{
5 public static int MinCharsToAppend(string s, string t)
6 {
7 int sLen = s.Length;
8 int tLen = t.Length;
9 int[,] next = new int[sLen + 1, 26];
10
11 for (int i = 0; i <= sLen; i++)
12 for (int j = 0; j < 26; j++)
13 next[i, j] = sLen;
14
15 int[] last = new int[26];
16 for (int i = 0; i < 26; i++) last[i] = sLen;
17
18 for (int i = sLen - 1; i >= 0; i--) {
19 last[s[i] - 'a'] = i;
20 for (int j = 0; j < 26; j++)
21 next[i, j] = last[j];
22 }
23
24 int pos = 0, appendCount = 0;
25 for (int i = 0; i < tLen; i++) {
26 if (next[pos, t[i] - 'a'] == sLen) {
27 appendCount += (tLen - i);
28 break;
29 }
30 pos = next[pos, t[i] - 'a'] + 1;
31 }
32
33 return appendCount;
34 }
35
36 public static void Main()
37 {
38 string s = "coaching";
39 string t = "coding";
40 Console.WriteLine(MinCharsToAppend(s, t));
41 }
42}The C# solution works by building the next table, precomputing where each character can next be found, and uses this table to determine the remaining characters in t that need appending into s to maintain order.