Validate a Bijective Substitution Cipher

Validate a Bijective Substitution Cipher

A substitution cipher replaces every occurrence of each character in an alphabet with another character. A substitution cipher is strictly bijective (one-to-one and onto) between two strings s1 and s2 if and only if:

1. Both strings have the exact same length.
2. Every character c in s1 maps consistently to the same corresponding character in s2 across all occurrences.
3. No two distinct characters in s1 map to the same character in s2.

Given two strings s1 and s2 consisting of printable ASCII characters (including letters, digits, spaces, and punctuation), write a function to determine if s1 and s2 define a valid bijective substitution cipher.

Function Signature

def is_bijective_cipher(s1: str, s2: str) -> bool:
    pass

Constraints

• 0 <= len(s1), len(s2) <= 10^5
• s1 and s2 contain only printable ASCII characters (ASCII codes 32 to 126 inclusive).
• Characters are case-sensitive ('A' and 'a' are distinct).

Examples

Example 1:

Input: s1 = "paper", s2 = "title"
Output: True
Explanation:
'p' -> 't'
'a' -> 'i'
'e' -> 'l'
'r' -> 'e'
Every character in s1 maps to a unique character in s2 consistently.

Example 2:

Input: s1 = "foo", s2 = "bar"
Output: False
Explanation:
'o' in s1 would need to map to both 'a' and 'r' in s2, which violates consistency.

Example 3:

Input: s1 = "ab", s2 = "aa"
Output: False
Explanation:
Both 'a' and 'b' map to 'a', violating the one-to-one requirement (collision).

Example 4:

Input: s1 = "Send $100!", s2 = "Wfoe %211?"
Output: True
Explanation:
Every character including spaces and symbols maps bijectively:
'S'->'W', 'e'->'f', 'n'->'o', 'd'->'e', ' '->' ', '$'->'%', '1'->'2', '0'->'1', '!'->'?'

Edge Cases to Consider

• Mismatched lengths: Immediately invalid.
• Empty strings: Two empty strings should return True.
• Fixed-points: A character mapping to itself (e.g., ' ' -> ' ') is valid as long as no other character also maps to ' '.