Find All Anagrams in a String TLE - algorithm

Question

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
Getting TEL for below code. What's wrong with my code?
import java.util.*;
public class FindAllAnagramsInAString_438{
public static void main(String[] args){
String s="abab";
String p="ab";
// String s="cbaebabacd";
// String p="abc";
System.out.println(findAnagrams(s,p));
}
public static List<Integer> findAnagrams(String s, String p) {
int i=0;
int j=p.length();
List<Integer> list=new ArrayList<>();
while(j<=s.length()){
//System.out.println("Substring >>"+s.substring(i,j));
if(isAnamgram(s.substring(i,j),p)){
list.add(i);
}
i++;
j++;
}
return list;
}
public static boolean isAnamgram(String s,String p){
HashMap<Character,Integer> map=new HashMap<>();
if(s.length()!=p.length()) return false;
for(int i=0;i<s.length();i++){
char chs=s.charAt(i);
char chp=p.charAt(i);
map.put(chs,map.getOrDefault(chs,0)+1);
map.put(chp,map.getOrDefault(chp,0)-1);
}
for(int val:map.values()){
if(val!=0) return false;
}
return true;
}
}

Answer 1

You shouldn't be creating a whole new frequency table for each substring. Given a table for index i, the table for index i+1 changes by only 2 counts. Just modify the previous table to get the next one. Also don't bother counting characters that don't appear in the search string.
Also, don't do the whole table comparison -- just keep track of the number of mismatched counts. Again this changes by at most 2 counts per index.
Finally, don't allocate a new substring for each index.

Answer 2

Looks good now
HashMap<Character,Integer> pMap=new HashMap<>();
HashMap<Character,Integer> sMap=new HashMap<>();
List<Integer> list=new ArrayList<>();
for(int i=0;i<p.length();i++){
char ch=p.charAt(i);
pMap.put(ch,pMap.getOrDefault(ch,0)+1);
}
int j=0;
while(j<p.length()){
char ch=s.charAt(j);
sMap.put(ch,sMap.getOrDefault(ch,0)+1);
j++;
}
int i=0;
while(j<s.length()){
if(pMap.equals(sMap)){
list.add(i);
}
sMap.put(s.charAt(j),sMap.getOrDefault(s.charAt(j),0)+1);
if((sMap.getOrDefault(s.charAt(i),0)-1)==0){
sMap.remove(s.charAt(i));
}else{
sMap.put(s.charAt(i),sMap.getOrDefault(s.charAt(i),0)-1);
}
i++;
j++;
}
return list;