Find All Anagrams in a String [Medium] — Fixed Window + Freq Match

Input: s="cbaebabacd", p="abc" → Output: [0,6]
Input: s="abab", p="ab"        → Output: [0,1,2]

from collections import Counter

def findAnagrams(s: str, p: str) -> list[int]:
    k = len(p)
    if len(s) < k: return []
    need = Counter(p)
    window = Counter(s[:k])
    matches = sum(1 for c in need if window[c] == need[c])
    res = [0] if matches == len(need) else []
    for i in range(k, len(s)):
        # Add right char
        c_in = s[i]; c_out = s[i-k]
        window[c_in] += 1
        if c_in in need:
            if window[c_in] == need[c_in]: matches += 1
            elif window[c_in] == need[c_in]+1: matches -= 1
        # Remove left char
        window[c_out] -= 1
        if c_out in need:
            if window[c_out] == need[c_out]: matches += 1
            elif window[c_out] == need[c_out]-1: matches -= 1
        if window[c_out] == 0: del window[c_out]
        if matches == len(need): res.append(i-k+1)
    return res

vector<int> findAnagrams(string s, string p) {
    if(s.size()<p.size()) return {};
    int freq[26]={}, win[26]={};
    for(char c:p) freq[c-'a']++;
    for(int i=0;i<p.size();i++) win[s[i]-'a']++;
    vector<int> res;
    auto eq=[&](){return equal(freq,freq+26,win);};
    if(eq()) res.push_back(0);
    for(int i=p.size();i<s.size();i++){
        win[s[i]-'a']++; win[s[i-p.size()]-'a']--;
        if(eq()) res.push_back(i-p.size()+1);
    }
    return res;
}