Find All Anagrams in a String
Given two strings $s$ and $p$, return an array of all the starting indices of $p$'s anagrams in $s$. The order of output does not matter.
Why Interviewers Ask This
Cisco evaluates candidates on their ability to optimize string processing for network packet analysis and pattern matching. This question tests mastery of the sliding window technique, a critical skill for handling real-time data streams efficiently without excessive memory overhead or nested loops that degrade performance.
How to Answer This Question
1. Clarify constraints: Ask about character sets (ASCII vs Unicode) and string lengths to determine if an O(N*M) brute force is acceptable or if O(N) is required. 2. Propose the Sliding Window approach: Explain how maintaining a frequency map of characters in 'p' allows you to slide a window across 's', updating counts incrementally rather than recalculating from scratch. 3. Define the update logic: Detail how adding a new character increments its count and removing the leftmost character decrements it, checking if the current window matches the target frequency map at each step. 4. Discuss edge cases: Mention empty strings, strings shorter than 'p', or inputs with no anagrams. 5. Analyze complexity: Explicitly state the time complexity as O(N) where N is the length of 's' and space complexity as O(1) since the alphabet size is fixed (26 lowercase letters), demonstrating the efficiency Cisco values for scalable systems.
Key Points to Cover
- Demonstrating the O(N) time complexity advantage over naive O(N*M) solutions
- Correctly implementing the sliding window update logic for both entry and exit characters
- Using a fixed-size frequency array to achieve O(1) auxiliary space complexity
- Explicitly validating the exact character match condition within the window
- Addressing edge cases like empty inputs or patterns longer than the source string
Sample Answer
To solve finding all anagrams of p in s efficiently, I would use the Sliding Window algorithm combined with a frequency array, which aligns well with Cisco's focus on high-performance network processing. First, I'd handle base cases where either string is null or p is longer than s, returning an empty list immediately. Next, I would initialize two arrays of size 26 to store character frequencies: one for the pattern p and another for the current window in s. As we iterate through s with a right pointer, we add the incoming character to our window count. Simultaneously, we maintain a left pointer; once the window size equals the length of p, we check if the current window's frequency array matches p's frequency array exactly. If they match, we record the starting index. Before moving the window forward, we decrement the count of the character leaving the window via the left pointer. This ensures we only traverse the string once, achieving O(N) time complexity. For example, if s is 'cbaebabacd' and p is 'abc', the algorithm identifies indices 0 ('cba') and 6 ('bac'). This approach avoids the O(N*M) cost of re-scanning substrings, making it ideal for large-scale data ingestion scenarios common in enterprise networking environments.
Common Mistakes to Avoid
- Recalculating the entire frequency map for every window position instead of incrementally updating it
- Failing to handle the case where the pattern string is longer than the source string
- Ignoring non-alphabetic characters or assuming a fixed ASCII set without verification
- Returning indices in a specific sorted order when the problem statement explicitly allows any order
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