The Question

String Transformation via Character Mapping

Given two strings str1 and str2 of the same length, determine if str1 can be transformed into str2 through a sequence of conversions. In a single conversion, you select one character occurring in str1 and replace all its occurrences with any other lowercase English letter. Return true if such a transformation is possible, otherwise return false. Note: A character can be transformed multiple times, but each step must convert all existing instances of the chosen character.

Java

HashMap

String Manipulation

Questions & Insights

Clarifying Questions

What is the maximum length of the strings? (Assuming up to

10^5

for a linear time solution).

Are the strings guaranteed to contain only lowercase English letters? (Assuming 'a'-'z').

Can a character be mapped to itself? (Yes, zero conversions are allowed).

Does the order of conversions matter? (Yes, if we map a -> b and then b -> c, all original as and bs become c. However, the problem asks if a sequence exists, which implies we can use intermediate temporary characters if available).

Assumptions:

Lengths are equal.

Only lowercase English letters.

If we need to swap characters (e.g., a -> b and b -> a), we need at least one free character in the alphabet not present in the target string str2 to act as a temporary buffer.

Thinking Process

Mapping Consistency: Each character in str1 must map to exactly one unique character in str2. If the same character at two different indices in str1 corresponds to different characters in str2, the transformation is impossible.

The "Cycle" Problem: Transformations can form chains (e.g., a -> b -> c). If these chains form a cycle (e.g., a -> b -> a), we need a temporary character not currently in the "occupied" set of str2 to break the cycle (e.g., a -> tmp, b -> a, tmp -> b).

Alphabet Exhaustion: If str2 contains all 26 characters of the lowercase English alphabet, and str1 is not already identical to str2, we cannot perform any transformation that requires a temporary buffer. More importantly, if all 26 characters are present in str2, no character is "free" to facilitate a transformation unless the strings are already identical.

Corner Case: If str1 equals str2, return true immediately (0 conversions).

Implementation Breakdown

Problem Set

Functional Requirements:

Validate that each

str1[i]

maps to a consistent

str2[i]

Ensure there is at least one character in the alphabet not present in str2 if transformations are required.

Constraints:

length(str1) == length(str2)

Alphabet size = 26.

Time Complexity:

O(N)

Space Complexity:

O(1)

(fixed size map/set).

Approach

Algorithm: One-to-One Mapping Validation + Alphabet Availability Check.

Data Structure: HashMap<Character, Character> or a fixed-size array for mapping, and a HashSet<Character> for tracking unique characters in str2.

Complexity: Time

O(N)

, Space

O(1)

(since the alphabet size is constant 26).

Implementation

Wrap Up

Advanced Topics

Readability vs Performance: While a HashMap is more readable, a fixed-size array int[26] is significantly faster and uses less memory for this specific alphabet constraint.

Cycle Detection: One might think about explicitly detecting cycles using Graph Theory (DFS/Tarjan's). However, in this specific problem, the simple existence of a "free" character is mathematically sufficient to resolve any number of cycles, making explicit cycle detection unnecessary.

Scaling to Unicode: If the character set was much larger (e.g., UTF-16), the uniqueCharsInStr2.size() < alphabetSize logic still holds, but the alphabet size would be much larger, making the "26 characters" constraint less likely to be hit.