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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.
1#include <stdio.h>
2#include <string.h>
3
4int minCharsToAppend(char* s, char* t) {
5 int sLen
The C solution declares a function minCharsToAppend that takes two strings, s and t, and calculates the minimum number of characters required to append to s to make t a subsequence. We use two indices to traverse the strings, incrementing sIndex for each iteration and only incrementing tIndex when characters match. Finally, we calculate the number of unmatched characters in t and return it.
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.
1import java.util.Arrays;
2
3public class MinCharsAppend {
4 public static int minCharsToAppend(String s, String t) {
5 int sLen = s.length(), tLen = t.length();
6 int[][] next = new int[sLen + 1][26];
7
8 for (int[] row : next) {
9 Arrays.fill(row, sLen);
10 }
11
12 int[] last = new int[26];
13 Arrays.fill(last, sLen);
14
15 for (int i = sLen - 1; i >= 0; i--) {
16 last[s.charAt(i) - 'a'] = i;
17 System.arraycopy(last, 0, next[i], 0, 26);
18 }
19
20 int pos = 0, appendCount = 0;
21 for (int i = 0; i < tLen; i++) {
22 if (next[pos][t.charAt(i) - 'a'] == sLen) {
23 appendCount += (tLen - i);
24 break;
25 }
26 pos = next[pos][t.charAt(i) - 'a'] + 1;
27 }
28
29 return appendCount;
30 }
31
32 public static void main(String[] args) {
33 String s = "coaching";
34 String t = "coding";
35 System.out.println(minCharsToAppend(s, t));
36 }
37}The Java solution involves creating a next array storing indices of upcoming occurrences of each character. It then strategically plans the minimal additions required by staying aware of precomputed locations.