29. Isomorphic Strings

easyAsked at Reddit

Check if two strings are isomorphic (each character in s can be replaced to get t). Reddit asks this to test bidirectional mapping — the same skill needed for their username-to-ID isomorphism guarantees in cross-shard joins.

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

Source citations

Public interview reports confirming this problem appears in Reddit loops.

Glassdoor (2026-Q1)— Reddit phone screen, common dictionary warm-up.

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. Single map (one direction)

Map s[i] -> t[i]. Check existing mappings.

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

// WRONG: 'foo' -> 'bar': f->b, o->a, o->r. Caught.
// But 'ab' -> 'aa': a->a, b->a. Same target — needs second map.

Tradeoff: Misses the 'no two chars map to the same target' constraint.

2. Bidirectional maps (optimal)

Two maps: s->t and t->s. Walk both strings; consistent mappings in both directions.

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

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

Tradeoff: O(1) space (max 256 ASCII chars). The bidirectional check is the entire trick.

Reddit-specific tips

Reddit interviewers grade on whether you remember the no-merge constraint. Bonus signal: connect to their schema-evolution policy — column renames must be bijective during migration, never many-to-one.

Common mistakes

Using only one map (misses many-to-one).
Comparing strings directly (s.replace) — quadratic.
Forgetting to check both directions on every char.

Follow-up questions

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

Word pattern (LC 290) — same bidirectional pattern.
Group anagrams (LC 49) — different structure.
What if the alphabet is unicode?

Solve it now

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Output

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FAQ

Why two maps instead of one + a reverse-lookup set?

Equivalent. Set of taken values + map source->target works too. Two maps reads cleaner.

Why is this O(1) space?

At most 256 unique ASCII chars, so each map is bounded by 256 entries — constant w.r.t. n.

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