Junior Python Developer Algorithm Interview

Approach: Use a sliding window with a character counter (Python's Counter or dictionary) to track character frequencies.

Time Complexity: O(n) where n is the length of string s

Space Complexity: O(k) where k is the size of the character set (usually 26 for lowercase letters)

Python Implementation:

from collections import Counter

def find_anagrams(s, p):
    result = []
    if len(p) > len(s):
        return result
        
    # Create character frequency counters
    p_counter = Counter(p)
    s_counter = Counter(s[:len(p)])
    
    # Check first window
    if p_counter == s_counter:
        result.append(0)
    
    # Slide the window
    for i in range(len(p), len(s)):
        # Add new character
        s_counter[s[i]] += 1
        # Remove oldest character
        s_counter[s[i - len(p)]] -= 1
        
        # Remove keys with zero count to keep counters comparable
        if s_counter[s[i - len(p)]] == 0:
            del s_counter[s[i - len(p)]]
        
        # Check if current window is an anagram
        if p_counter == s_counter:
            result.append(i - len(p) + 1)
    
    return result

Key Python-Specific Points:

• Using collections.Counter for efficient character counting
• Direct comparison of Counter objects (p_counter == s_counter)
• Removing zero-count entries to ensure proper comparison
• Sliding window implementation with constant space complexity