Solution Explanation: Isomorphic Strings

The problem asks to determine if two strings, s and t, are isomorphic. Two strings are isomorphic if there's a one-to-one mapping between their characters such that replacing each character in s with its corresponding character in t results in an identical string. Crucially, this mapping must be consistent; no character in s can map to multiple characters in t.

Approach: Using Hash Maps (or Arrays)

The most efficient approach uses two hash maps (or arrays, for better performance if the character set is known and limited, as in this case - ASCII characters). One map (d1) tracks the mapping from characters in s to characters in t, and the other (d2) tracks the reverse mapping.

Algorithm:

1. Initialization: Create two hash maps (or arrays of size 256 for ASCII characters), d1 and d2, initialized to empty or 0.

2. Iteration: Iterate through both strings s and t simultaneously using zip (Python) or indexing (other languages).

3. Mapping Check: For each pair of characters (a, b) from s and t, respectively:

   • Check if a is already mapped in d1. If it is, verify that its mapping is b. If not, return False (violates one-to-one mapping).
   • Similarly, check if b is already mapped in d2. If it is, verify that its mapping is a. If not, return False.
   • If both checks pass, update the mappings in both d1 and d2: d1[a] = b and d2[b] = a.

4. Return True: If the loop completes without returning False, it means all characters have consistent mappings, and the strings are isomorphic. Return True.

Time and Space Complexity

• Time Complexity: O(n), where n is the length of the strings. We iterate through the strings once.
• Space Complexity: O(C), where C is the size of the character set. In this case, C is 256 for ASCII characters (using arrays); using hash maps, the space complexity depends on the number of unique characters. Using arrays is generally faster than hash maps in this scenario because array access is O(1) whereas hash map lookups can be O(1) on average but O(n) in the worst case.

Code Examples (Multiple Languages)

The provided code examples demonstrate this approach using different programming languages. Note that the array-based solutions are generally more efficient than the hash map-based solutions for this problem.

Python (Array-based):

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        d1, d2 = [0] * 256, [0] * 256
        for i, (a, b) in enumerate(zip(s, t), 1):
            a, b = ord(a), ord(b)
            if d1[a] != d2[b]:
                return False
            d1[a] = d2[b] = i
        return True

Java (Array-based):

class Solution {
    public boolean isIsomorphic(String s, String t) {
        int[] d1 = new int[256];
        int[] d2 = new int[256];
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            char a = s.charAt(i), b = t.charAt(i);
            if (d1[a] != d2[b]) {
                return false;
            }
            d1[a] = i + 1;
            d2[b] = i + 1;
        }
        return true;
    }
}

(Other language examples are similar in structure, using arrays or hash maps as appropriate.)

This detailed explanation, along with the provided code examples, should give a comprehensive understanding of how to solve the "Isomorphic Strings" problem efficiently. The array-based solutions offer a slight performance advantage over hash map solutions in this specific problem due to the fixed and limited size of the character set.