Teaching Kids Programming – Is Subsequence Algorithm via Recursion (Greedy)

Teaching Kids Programming: Videos on Data Structures and Algorithms

Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., “ace” is a subsequence of “abcde” while “aec” is not).

Example 1:
Input: s = “abc”, t = “ahbgdc”
Output: true

Example 2:
Input: s = “axc”, t = “ahbgdc”
Output: false

Constraints:
0 <= s.length <= 100
0 <= t.length <= 104
s and t consist only of lowercase English letters.

Follow up: Suppose there are lots of incoming s, say s1, s2, …, sk where k >= 10^9, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

Is Subsequence Algorithm via Two Pointer

We have learned the Two Pointer Algorithm (Teaching Kids Programming – Is Subsequence Algorithm via Two Pointer) to check if a string is a subsequence of another string.

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s) > len(t):
            return False
        i = 0
        j = 0
        while i < len(s) and j < len(t):
            if s[i] == t[j]:
                i += 1
                j += 1
            else:
                j += 1
        return i == len(s)

The time complexity is O(N) where N is the length of the longer string t.

Is Subsequence Algorithm via Recursion + Greedy

We can define a Recursive Function takes take two parameters, the i and j indicating the current position for s and t respectively. Then we have two cases: if s[i] == s[t] or s[i] != s[t]. When we have a match, we can advance two pointers. Otherwise, we have to move the j pointer to next.

The idea is same as above two pointer approach, however implemented in Recursion.

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s) > len(t):
            return False        

        def dfs(i, j):
            if i == len(s):
                return True
            if j == len(t):
                return False
            if s[i] == t[j]:
                return dfs(i + 1, j + 1)
            return dfs(i, j + 1)
        return dfs(0, 0)

We can check from the end backwards as well:

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s) > len(t):
            return False        

        def dfs(i, j):
            if i < 0:
                return True
            if j < 0:
                return False
            if s[i] == t[j]:
                return dfs(i - 1, j - 1)
            return dfs(i, j - 1)
        
        return dfs(len(s) - 1, len(t) - 1)

The time complexity is O(N) where N is the length for the longer string t.

String Subsequence Algorithms:

• Teaching Kids Programming – Is Subsequence Algorithm via Recursion (Greedy)
• Teaching Kids Programming – Is Subsequence Algorithm via Two Pointer
• The Subsequence Algorithm for Two Strings – How to Check if a String is Subsequence of Another?
• GoLang Function of Checking Is Subsequence
• Algorithms to Check if a String is a Subsequence String of Another String
• in Python, you can do this in Pythonic way: Python Function to Check Subsequence

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