Isomorphic Strings Leetcode Problem 205 [Python Solution]

In this blog post, we're going to tackle the Isomorphic Strings problem, a LeetCode question with a difficulty level marked as "Easy." This problem falls under the category of Arrays & Hashing and has been associated with companies like Meta.

If you want to dive deeper into this problem, you can find the question on LeetCode here.

Problem Overview

The problem statement is relatively straightforward: given two strings, s and t, our goal is to determine whether they are isomorphic.

Two strings are considered isomorphic if the characters in one string can be replaced to get the other string, subject to a few constraints.

Here's the definition of isomorphic strings in this context:
– All occurrences of a character in one string must be replaced with another character in the other string.
– The order of characters must be preserved during this replacement.
– No two characters in one string may map to the same character in the other, but a character can map to itself.

Let's look at a couple of examples to illustrate this concept:

Example 1:

Input: s = "egg", t = "add"
Output: true

In this example, we can see that "e" in s maps to "a" in t, and "g" in s maps to "d" in t.

All characters have unique mappings, and the order is preserved, so these strings are isomorphic.

Example 2:

Input: s = "foo", t = "bar"
Output: false

In this case, "f" in s maps to "b" in t", but "o" insalso maps to "o" int`.

The mapping is not unique, and the order is not preserved, so these strings are not isomorphic.

Constraints:

• 1 <= s.length <= 5 * 10^4
• t.length == s.length
• s and t consist of any valid ASCII character.

Now that we understand the problem statement, let's delve into how we can efficiently solve it with Python.

Efficient Python Code Solution

Here's an efficient Python solution to the Isomorphic Strings problem:

def isIsomorphic(self, s: str, t: str) -> bool:
    mapST, mapTS = {}, {}

    for c1, c2 in zip(s, t):
        if (c1 in mapST and mapST[c1] != c2) or (c2 in mapTS and mapTS[c2] != c1):
            return False
        mapST[c1] = c2
        mapTS[c2] = c1

    return True

This code uses two dictionaries, mapST and mapTS, to store the mappings of characters from s to t and vice versa.

It iterates through the characters of both strings and checks whether the current character has a consistent mapping.

If at any point a character violates the isomorphic condition (either it already has a different mapping or it maps to two different characters), the function returns False.

If no violations are found during the iteration, it returns True, indicating that the strings are isomorphic.

The time and space complexity of this solution are both O(n), where n is the length of the input strings.

This is because we iterate through both strings once, and the hash maps used for mapping characters also have a constant-time lookup and insertion, making the overall solution efficient.

Reasoning Behind Our Approach

The key idea behind this efficient approach is to use two hash maps (mapST and mapTS) to track the mappings of characters from s to t and from t to s.

By checking and updating these mappings while iterating through the strings, we ensure that each character in one string corresponds to a unique character in the other, preserving order.

If at any point we find a character that violates this rule, we return False.

The use of hash maps allows for constant-time lookups and insertions, ensuring the overall efficiency of the solution.

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Conclusion

In this blog post, we've explored the Isomorphic Strings problem, a LeetCode question that challenges us to determine if two given strings are isomorphic.

We've discussed the problem's definition and constraints and provided an efficient Python solution using hash maps.

The solution efficiently checks for isomorphism, and its time and space complexity is O(n), making it a practical and effective way to tackle this problem.

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