Is Subsequence
Given two strings $s$ and $t$, return true if $s$ is a subsequence of $t$, or false otherwise.
Why Interviewers Ask This
Cisco evaluates this question to assess a candidate's ability to implement efficient two-pointer logic without unnecessary overhead. It tests fundamental algorithmic thinking, specifically the capacity to solve problems with O(n) time complexity and O(1) space constraints. The interviewer looks for clarity in handling edge cases like empty strings and the ability to translate logical conditions into clean, bug-free code quickly.
How to Answer This Question
1. Clarify the definition: Briefly restate that a subsequence maintains relative order but allows skipping characters, unlike a substring which must be contiguous.
2. Identify the optimal strategy: Immediately propose the Two-Pointer technique as the most efficient solution, noting it avoids recursion or dynamic programming overhead.
3. Walk through the logic: Explain initializing one pointer for string s and another for t. Describe iterating through t while advancing the s pointer only when characters match.
4. Define termination conditions: Explicitly state that if the s pointer reaches the end of s, return true; if t ends before s is exhausted, return false.
5. Address edge cases: Mention handling scenarios where s is empty (always true) or longer than t (always false).
6. Analyze complexity: Conclude by confirming the time complexity is O(m) where m is the length of t, and space complexity is O(1).
Key Points to Cover
- Demonstrating the Two-Pointer pattern as the standard solution for subsequence checks
- Explicitly stating O(n) time and O(1) space complexity analysis
- Handling the specific edge case where the subsequence string is empty
- Maintaining relative order logic rather than checking for contiguous substrings
- Providing a clear, step-by-step walkthrough of the pointer movement logic
Sample Answer
To determine if string s is a subsequence of string t, I would use the Two-Pointer approach, which is optimal for this problem. First, I'd check for edge cases: if s is empty, it is trivially a subsequence of any string, so I return true. If s is longer than t, it cannot be a subsequence, so I return false.
Next, I initialize two pointers, let's call them i for s and j for t, both starting at index zero. I will iterate through string t using pointer j. For each character in t, I compare it with the current character in s pointed to by i. If they match, it means we found the next required character for our subsequence, so I increment i to move to the next character in s. Regardless of whether there was a match, I always increment j to continue scanning t.
The loop continues until either j reaches the end of t or i reaches the end of s. If i successfully reaches the length of s, it means every character in s was found in t in the correct relative order, so I return true. If the loop finishes and i has not reached the end of s, then s is not a subsequence, and I return false.
This approach runs in O(n) time where n is the length of t, and uses O(1) extra space, making it highly efficient for large inputs typical in network packet processing scenarios Cisco handles.
Common Mistakes to Avoid
- Confusing subsequence with substring by requiring characters to be contiguous
- Using nested loops which results in inefficient O(n*m) time complexity
- Failing to handle the case where the subsequence string is empty
- Not analyzing the time and space complexity after writing the solution
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