find all anagrams in a string leetcode

class Solution {
public:
    // Time Complexity:- O(max(N,26))
    // Space Complexity:- O(26)
    vector<int> findAnagrams(string s, string p) {
        int n = s.length(),m = p.length();

        // not possible for this case
        if(m>n){
            return {};
        }

        // if p is an anagram of some substring of s, then their frequency must be same
        vector<int> demand_frequency(26),substring_frequency(26),ans;
        for(auto& c:p){
            demand_frequency[c-'a']++;
        }
        for(int i=0;i<m;i++){
            substring_frequency[s[i]-'a']++;
        }

        // deviation = number of characters whose frequency aren't equal to demand frequency
        int deviation = 0;
        for(int i=0;i<26;i++){
            deviation += (demand_frequency[i]!=substring_frequency[i]);
        }

        // fill answer for first substring of m length
        if(deviation==0){
            ans.push_back(0);
        }

        for(int i=m;i<n;i++){
            int j = s[i] - 'a'; // add the current character to current frequency array
            substring_frequency[j]++;
            // if their frequency matches now, it means previously their frequency didn't match, hence we do deviation-- in this case
            if(substring_frequency[j]==demand_frequency[j]){
                deviation--;
            }
            // if current frequency of character exceeds the demand frequency of character by exactly 1, means previously they have same frequency
            else if(substring_frequency[j]-demand_frequency[j]==1){
                deviation++;
            }

            j = s[i-m] - 'a';
            substring_frequency[j]--;
            // if their frequency matches now, it means previously their frequency didn't match, hence we do deviation-- in this case
            if(substring_frequency[j]==demand_frequency[j]){
                deviation--;
            }
            // if demand frequency of character exceeds the current frequency of character by exactly 1, means previously they have same frequency
            else if(demand_frequency[j]-substring_frequency[j]==1){
                deviation++;
            }

            // if deviation is 0, means all characters have their desired frequency
            if(deviation==0){
                ans.push_back(i-m+1);
            }
        }

        return ans;
    }
};