27. Isomorphic Strings

easyAsked at Plaid

Determine if two strings are isomorphic — one character maps to one other character consistently. Plaid asks this because tokenizing merchant strings under a consistent mapping is the same primitive.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Plaid loops.

LeetCode Discuss (2026)— Plaid OA — frame as merchant token mapping.
Glassdoor (2025)— Plaid intro.

Problem

Given two strings s and t, determine if they are isomorphic. Two strings s and t are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character, but a character may map to itself.

Constraints

1 <= s.length <= 5 * 10^4
t.length == s.length
s and t consist of any valid ASCII character.

Examples

Example 1

Input

s = "egg", t = "add"

Output

true

Example 2

Input

s = "foo", t = "bar"

Output

false

Example 3

Input

s = "paper", t = "title"

Output

true

Approaches

1. One-way mapping

Map each s[i] to t[i]. Fail if a conflicting mapping appears.

Time: O(n)
Space: O(1) for ASCII

function isIsomorphic(s, t) {
  const m = new Map();
  for (let i = 0; i < s.length; i++) {
    if (m.has(s[i])) {
      if (m.get(s[i]) !== t[i]) return false;
    } else m.set(s[i], t[i]);
  }
  return true;
}

Tradeoff: Misses the 'no two chars map to the same target' rule. Fails on s="ab", t="aa".

2. Bidirectional mapping

Track s->t AND t->s. Both must be consistent.

Time: O(n)
Space: O(1) for ASCII

function isIsomorphic(s, t) {
  const st = new Map(), ts = new Map();
  for (let i = 0; i < s.length; i++) {
    if (st.has(s[i]) && st.get(s[i]) !== t[i]) return false;
    if (ts.has(t[i]) && ts.get(t[i]) !== s[i]) return false;
    st.set(s[i], t[i]);
    ts.set(t[i], s[i]);
  }
  return true;
}

Tradeoff: Bijective mapping ensured. The 'two maps in parallel' is the standard pattern for one-to-one matching.

Plaid-specific tips

Plaid grades this on whether you spot the bijection requirement. Bonus signal: name it 'bijective' aloud. Connect it to merchant-tokenization — where 'STARBUCKS #234' and 'STARBUCKS #567' must reduce to the same canonical token, but two different merchants must never collapse to the same token.

Common mistakes

One-way mapping only — misses 'two source chars to one target' bug.
Using a single map keyed by 'char-pair' string — wasteful allocation.
Comparing lengths first then iterating — fine but the problem guarantees equal length.

Follow-up questions

An interviewer at Plaid may pivot to one of these next:

Word pattern (LC 290) — same idea, words instead of chars.
Anagram check across multiple strings.
Group anagrams (LC 49).

Solve it now

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Output

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FAQ

Why bidirectional?

The rule says no two source chars map to the same target. Tracking only s->t doesn't catch that case.

Why two maps instead of one combined?

Two maps make the bijection invariant readable. A combined char-pair key works but is harder to reason about.

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