🦄 Isomorphic Runes: One-to-One Mapping

Difficulty: Beginner Tags: hashmap, strings, bijection, pattern-matching Series: CS 102: Problem Solving & Logic

Problem

In the mystical world of runes, two inscriptions are considered isomorphic if one can be transformed into the other through a consistent one-to-one character mapping.

Given two strings s and t, determine if they are isomorphic. Strings are isomorphic if characters in s can be replaced to get t, where: - Each character in s maps to exactly one character in t - No two characters in s map to the same character in t - The mapping must be bijective (one-to-one in both directions)

Real-World Application

Isomorphic string checking appears in: - Data deduplication - identifying duplicate records with different formats - Pattern matching - finding similar structures in text or code - Schema mapping - aligning database schemas or API formats - Word games - pattern-based word puzzles - Compiler design - variable renaming and canonicalization

Input

data = {
    's': str,  # First string
    't': str   # Second string (same length as s)
}

Output

bool  # True if isomorphic, False otherwise

Constraints

• |s| = |t| (strings have equal length)
• |s|, |t| ≤ 100,000
• Strings contain ASCII characters

Examples

Example 1: Basic Isomorphic Strings

Input: {'s': 'egg', 't': 'add'}

Output: True

Explanation: - 'e' → 'a' (first char maps) - 'g' → 'd' (second char maps) - 'g' → 'd' (third char must map same as second) - Mapping: {'e': 'a', 'g': 'd'} ✓ - No conflicts, bijection maintained

Example 2: Non-Isomorphic Strings

Input: {'s': 'foo', 't': 'bar'}

Output: False

Explanation: - 'f' → 'b' (first char maps) - 'o' → 'a' (second char maps) - 'o' → 'r' (CONFLICT! 'o' already mapped to 'a') - The mapping is inconsistent

Example 3: Bijection Violation

Input: {'s': 'badc', 't': 'baba'}

Output: False

Explanation: - 'b' → 'b', 'a' → 'a', 'd' → 'b', 'c' → 'a' - Two different characters ('b' and 'd') both map to 'b' → Not one-to-one! - Two different characters ('a' and 'c') both map to 'a' → Not one-to-one!

Example 4: Same Character Repeated

Input: {'s': 'paper', 't': 'title'}

Output: True

Explanation: - 'p' → 't', 'a' → 'i', 'p' → 't' (consistent!), 'e' → 'l', 'r' → 'e' - Mapping: {'p': 't', 'a': 'i', 'e': 'l', 'r': 'e'} ✓ - Reverse mapping also valid: {'t': 'p', 'i': 'a', 'l': 'e', 'e': 'r'} ✓

What You'll Learn

• How to maintain bidirectional (bijective) mappings using two hash maps
• When to use early exit strategies to optimize performance
• Pattern recognition in string transformations
• The difference between mapping and bijection

Why This Matters

Technical interviews frequently test your ability to recognize when bijection is required vs simple mapping, and how to efficiently track both directions of a relationship.

Starter Code

def challenge_function(data):
    """
    Determine if two strings are isomorphic.

    Args:
        data: dict with keys 's' (str) and 't' (str)

    Returns:
        bool: True if s and t are isomorphic, False otherwise
    """
    # Your implementation here
    pass