Is Subsequence

class Solution {

    public boolean sequenceMatcher(String str1, String str2) {

        Integer len1 = str1.length(), len2 = str2.length();
        if (len1 == 0)
            return true;

        int[][] matchTable = new int[len1 + 1][len2 + 1];
        for (int j = 1; j <= len2; ++j) {
            for (int i = 1; i <= len1; ++i) {
                if (str1.charAt(i - 1) == str2.charAt(j - 1))
                    matchTable[i][j] = matchTable[i - 1][j - 1] + 1;
                else
                    matchTable[i][j] = Math.max(matchTable[i][j - 1], matchTable[i - 1][j]);
            }
            if (matchTable[len1][j] == len1)
                return true;
        }
        
        return false;
    }
}

The provided Java solution addresses the problem of determining whether one string (str1) is a subsequence of another string (str2). The method sequenceMatcher accomplishes this by employing a dynamic programming approach using a 2D array, matchTable, to store the lengths of matching subsequences found during the iteration through both strings.

Here's how the solution works:

• Initiate matchTable where matchTable[i][j] represents the length of the longest subsequence common to str1 up to i and str2 up to j.
• Iterate through each character in str2 (outer loop) and for each character of str2, iterate through each character of str1 (inner loop).
• If characters from both strings match, the value in the table is derived from the top-left diagonal cell incremented by one. This indicates the continuation of an existing subsequence.
• If there's no match, the cell's value is the maximum of either the cell directly above or the cell to the left, representing the longest subsequence found so far without including the current character.
• After updating the table through each character of str2, check if the subsequence having the length of str1 has been found by looking at the last cell of the current column of str2.
• If str1 is indeed a subsequence at any point during the traversal of str2, return true.
• If after completing the traversal, no such subsequence is identified, return false.

This method efficiently determines the relationship between the two strings, optimizing checks and balances through dynamic programming, and provides an immediate exit upon confirmation that a subsequence exists, making it computationally efficient for varying lengths of input strings.

class Solution:
    def isSubsequence(self, sub: str, main: str) -> bool:
        len_sub, len_main = len(sub), len(main)

        if len_sub == 0:
            return True

        dp_table = [[0] * (len_main + 1) for _ in range(len_sub + 1)]

        for i in range(1, len_main + 1):
            for j in range(1, len_sub + 1):
                if sub[j - 1] == main[i - 1]:
                    dp_table[j][i] = dp_table[j - 1][i - 1] + 1
                else:
                    dp_table[j][i] = max(dp_table[j][i - 1], dp_table[j - 1][i])

            if dp_table[len_sub][i] == len_sub:
                return True

        return False

The solution provided in Python addresses the problem of checking if one string is a subsequence of another. The function isSubsequence takes two strings, sub and main, and determines if sub is a subsequence within main.

The initial check in the solution handles the case where the subsequence is empty, immediately returning True since an empty sequence is trivially a subsequence of any string.

A dynamic programming approach is used, establishing a 2D list dp_table where dp_table[j][i] will hold the length of longest common subsequence of sub[0:j] and main[0:i].

• A nested loop iterates through both sub and main.
  • If characters from sub and main match, the value in the DP table is updated to carry over the length from the previous matched position incremented by one.
  • If no match occurs, the function carries forward the maximum subsequence length found so far either from skipping a character in sub or main.

The function concludes if the value at dp_table[len_sub][i] is equal to len_sub at any column i, signifying all characters of sub have been matched in sequence within main.

If the end of main is reached and sub has not been fully matched, False is returned. This approach efficiently checks the subsequence condition with the help of dynamic programming, ensuring all possible subsequences of main are explored for a potential complete match with sub.