242: Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nTo solve this problem, we can use a dictionary to count the frequency of 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nTo solve this problem, we can use a dictionary to count the frequency of each character in both strings. Then, we can compare the two dictionaries to see if they have the same character frequencies. If they do, then the strings are anagrams.\n\n# Complexity\n- Time complexity:\nO(n), where n is the length of the input strings. We loop through each character in both strings once.\n\n- Space complexity:\nO(1), since the size of the dictionaries is constant (26 lowercase English letters). However, if the input strings contain Unicode characters, the space complexity would be O(k), where k is the number of unique Unicode characters in the input strings.\n\n# Code\n```\nclass Solution:\n  def isAnagram(self, s: str, t: str) -> bool:\n    # Check if lengths of s and t are equal\n    if len(s) != len(t):\n        return False\n    \n    # Create two dictionaries to store the frequency of each character in s and t\n    s_freq = {}\n    t_freq = {}\n    \n    # Loop through each character in s and t\n    for i in range(len(s)):\n        # Increment the frequency of the current character in the corresponding dictionary\n        s_freq[s[i]] = s_freq.get(s[i], 0) + 1\n        t_freq[t[i]] = t_freq.get(t[i], 0) + 1\n        \n    # Compare the two dictionaries\n    return s_freq == t_freq\n\n```
['Hash Table', 'String', 'Sorting', 'Python', 'Python3']