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Solution 15: Sliding Permutation Window (Find All Anagrams in a String)

Approach Explanation

We use a sliding window of size len(p) and compare the character frequencies of the window with the frequencies of p.

Step-by-Step Logic

  1. Count frequencies of p in pCount map.
  2. Maintain sCount for the current window in s.
  3. As we slide the window:
    • Add the new char to sCount.
    • Remove the old char (leftmost) from sCount.
    • If sCount == pCount, add start index to result.

Complexity

  • Time Complexity: O(N).
  • Space Complexity: O(1) (alphabet size).

Code

from collections import Counter

def find_anagrams(s, p):
    ns, np = len(s), len(p)
    if ns < np: return []
    
    p_count = Counter(p)
    s_count = Counter(s[:np])
    
    res = []
    if s_count == p_count:
        res.append(0)
        
    for i in range(1, ns - np + 1):
        # Update window
        prev_char = s[i-1]
        new_char = s[i + np - 1]
        
        s_count[new_char] += 1
        s_count[prev_char] -= 1
        
        if s_count[prev_char] == 0:
            del s_count[prev_char]
            
        if s_count == p_count:
            res.append(i)
            
    return res