# Solving Leetcode Interviews in Seconds with AI: Isomorphic Strings


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "205" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given two strings s and t, determine if they are isomorphic. Two strings s and t are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.   Example 1:  Input: s = "egg", t = "add" Output: true Explanation: The strings s and t can be made identical by:  Mapping 'e' to 'a'. Mapping 'g' to 'd'.   Example 2:  Input: s = "foo", t = "bar" Output: false Explanation: The strings s and t can not be made identical as 'o' needs to be mapped to both 'a' and 'r'.  Example 3:  Input: s = "paper", t = "title" Output: true    Constraints:  1 <= s.length <= 5 * 104 t.length == s.length s and t consist of any valid ascii character.  

	# Explanation
	Here's a solution to determine if two strings are isomorphic, focusing on efficiency and clarity:

*   **Core Idea:** Use two dictionaries (hash maps) to track the mapping between characters in `s` and `t`. One dictionary maps characters from `s` to `t`, and the other maps characters from `t` to `s`. This ensures that the mapping is bijective (one-to-one).
*   **Verification:** Iterate through the strings `s` and `t` simultaneously. For each character pair `s[i]` and `t[i]`, check if the mappings in the dictionaries are consistent. If a mapping conflict is found, the strings are not isomorphic.
*   **Early Exit:** If the lengths of the strings are different, they cannot be isomorphic.

*   **Complexity:** O(n) runtime, O(1) storage (since the character set is bounded by ASCII, the dictionaries can grow to a maximum fixed size).

	
	# Code
	```python
	def isIsomorphic(s: str, t: str) -> bool:
    """
    Determines if two strings are isomorphic.

    Args:
        s: The first string.
        t: The second string.

    Returns:
        True if the strings are isomorphic, False otherwise.
    """

    if len(s) != len(t):
        return False

    s_to_t = {}
    t_to_s = {}

    for i in range(len(s)):
        char_s = s[i]
        char_t = t[i]

        if char_s in s_to_t:
            if s_to_t[char_s] != char_t:
                return False
        else:
            s_to_t[char_s] = char_t

        if char_t in t_to_s:
            if t_to_s[char_t] != char_s:
                return False
        else:
            t_to_s[char_t] = char_s

    return True
	```