Solution: Problem Challenge 2: String Anagrams

import java.util.*;

class Solution {
    public List<Integer> findStringAnagrams(String str, String pattern) {
      List<Integer> resultIndices = new ArrayList<Integer>();

      Map<Character, Integer> patternMap = new HashMap<>();
      for(Character c: pattern.toCharArray()) {
        patternMap.put(c, patternMap.getOrDefault(c, 0)+1);
      }

      char[] arr = str.toCharArray();
      int n = arr.length, plen = pattern.length();
      Map<Character, Integer> map = new HashMap<>();
      for(int i = 0; i < plen; i ++) {
        map.put(arr[i], map.getOrDefault(arr[i], 0)+1);
      }

      if(isMatch(map, patternMap)) {
        resultIndices.add(0);
      }

      int st = 0;
      for(int end = plen; end < n; end++) {
        map.put(arr[end], map.getOrDefault(arr[end], 0)

import math
from collections import Counter

class Solution:
  def findStringAnagrams(self, text, pattern):
    result = []
    if text == "":
      return result 

    # pattern frequency count
    pattern_counter = Counter(pattern)

    # initial window frequency count
    text_counter = {}
    for i in range(len(pattern)):
      text_counter[text[i]] = text_counter.get(text[i], 0)
      text_counter[text[i]] += 1

    i = len(pattern) - 1
    j = 0
    
    while i < len(text):
      # if counters match we have an anagram
      if text_counter == pattern_counter:
        result.append(j)
      
      # move i then add the i frequency to the map
      i += 1
      if i == len(text):
        break
      text_counter[text[i]] = text_counter.get(text[i], 0)
      text_counter[