easyHashingPattern: Hash Map / Bijection
Isomorphic String Mapping Solution
Problem Statement
Given two strings `s` and `t`, determine if there exists a one-to-one mapping between the characters of `s` and `t`. The mapping is valid if every character in `s` maps to a unique character in `t` and vice versa.
Examples
Example 1:
Input:{"s":"egg","t":"add"}
Output:true
Example 2:
Input:{"s":"foo","t":"bar"}
Output:false
Constraints
- 1 <= s.length <= 5 * 10^4
- t.length == s.length
Time: O(N) Space: O(1)
Use two arrays of size 256 for mapping s->t and t->s. If mismatch found, return false. Time O(N), Space O(1).
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