Append Characters to String to Make Subsequence - Solution & Explanation

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.

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.

buildNext function precomputes the next occurrences of each character for every index. Using this information, we find out how to move through s to match t and calculate how many characters in t are not matched, hence needing to be appended.