Count-Occurences-of-Anagrams

Problem Link : https://www.geeksforgeeks.org/anagram-substring-search-search-permutations/
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Complexity : Time - O(N), Memory - O(26)
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int search(String pat, String txt) {
        int cnt=0, i=0, j=0;
        int[] freqp=new int[26], freqt=new int[26];
        boolean f=false;
        
        for(char ch : pat.toCharArray())
            ++freqp[ch-'a'];
        while(i<txt.length()){
            char ch=txt.charAt(i);
            
            if(freqp[ch-'a']==0){//if the curr char not exist in pat then reset the txt_freq[]
                j=++i;
                freqt=new int[26];
                f=false;
            } else if(freqp[ch-'a']==freqt[ch-'a']){//if the curr char == pat_freq then decrese the window
                --freqt[txt.charAt(j)-'a'];
                ++j;
            }else{////if the txt_freq[curr_char]<pat_freq[curr_char] then increase txt_freq[curr_char] as it is a potential candidate
                ++freqt[ch-'a'];
                ++i;
            }
            
            if(i-j==pat.length()){
                ++cnt;
            }
        }   
        
        return cnt;
    }