28. Isomorphic Strings

easyAsked at Datadog

Determine if two strings are isomorphic — every character in s maps to one in t under a consistent 1-to-1 substitution. Datadog asks this for the two-way-mapping insight — same shape as their tag-name canonicalization on ingest.

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

Source citations

Public interview reports confirming this problem appears in Datadog loops.

Glassdoor (2026-Q1)— Datadog phone screen string problem.

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 map (broken)

Single map s_char -> t_char. Misses the 'no two s chars map to the same t char' rule.

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

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: Wrong on cases like 'foo' -> 'bar' (passes incorrectly). Datadog will catch this.

2. Two-way mapping (optimal)

Track both s->t and t->s. Both must be consistent (bijection).

Time: O(n)
Space: O(1) bounded by alphabet

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

Tradeoff: Two maps enforce bijection. Datadog-canonical — directly maps to their tag-canonicalization integrity check.

Datadog-specific tips

Datadog will probe with the 'foo' -> 'bar' counterexample to see if you spot that one-way mapping fails. Articulate the bijection requirement BEFORE coding — the two maps are non-obvious unless you've thought it through.

Common mistakes

Single-direction map — passes 'egg'/'add' but fails 'foo'/'bar' or 'badc'/'baba'.
Using Object.create(null) without thinking about prototype pollution — Map is the safer default.
Comparing first-occurrence positions instead of using maps — works (encode-by-index) but allocates more.

Follow-up questions

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

Word Pattern (LC 290) — same idea but pattern is char per word.
Group Anagrams (LC 49) — different kind of equivalence relation.
Datadog-style: canonicalize tag IDs across two ingestion paths.

Solve it now

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Output

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FAQ

Why two maps?

Isomorphism requires a BIJECTION between character sets. One map allows two distinct s-chars to map to the same t-char, violating the 1-to-1 rule.

Could you encode characters by first-occurrence index?

Yes — transform both strings into 'position of first occurrence' arrays and compare. Equivalent but allocates O(n) per string.

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