How do you solve Is Subsequence on LeetCode?

Is Subsequence (LeetCode #392) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using two pointers can significantly reduce time complexity.

What is the brute force solution for Is Subsequence?

The brute force approach for Is Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks all possible subsequences of string 't' to see if 's' matches any of them. This method is straightforward but inefficient as it can involve a lot of unnecessary checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Is Subsequence?

Is Subsequence uses the following concepts: Two Pointers, String, Dynamic Programming. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Is Subsequence?

Clarify the definition of subsequence with the interviewer. Discuss both brute force and optimal solutions during the interview. Practice explaining your thought process clearly and concisely.

What are common mistakes when solving Is Subsequence?

Not considering edge cases like empty strings. Confusing subsequences with substrings.

#392

Is Subsequence

Easy

Two PointersStringDynamic ProgrammingTwo PointersSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Using two pointers, we can efficiently check if 's' is a subsequence of 't' by iterating through both strings simultaneously. This method is much faster as it only requires a single pass through both strings.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, one for 's' and one for 't'.
2Step 2: Traverse through 't' using its pointer.
3Step 3: If characters at both pointers match, move the pointer for 's' forward.
4Step 4: Always move the pointer for 't' forward.
5Step 5: If the pointer for 's' reaches the end of 's', return true; otherwise, return false.

solution.py8 lines

1# Full working Python code
2def isSubsequence(s, t):
3    s_ptr, t_ptr = 0, 0
4    while s_ptr < len(s) and t_ptr < len(t):
5        if s[s_ptr] == t[t_ptr]:
6            s_ptr += 1
7        t_ptr += 1
8    return s_ptr == len(s)

ℹ

Complexity note: The time complexity is O(n) because we traverse each string at most once, making it linear in relation to the length of 't'. The space complexity is O(1) as we only use a few variables.

1Using two pointers can significantly reduce time complexity.
2Understanding subsequences helps in recognizing patterns in string problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.