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28. Isomorphic Strings

easyAsked at Vercel

Determine if two strings are isomorphic (consistent one-to-one character mapping). Vercel asks this because the bidirectional-mapping check is the same shape as their canary deploy 'this version maps consistently to that traffic slice' validator.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Vercel loops.

  • Glassdoor (2025-12)— Vercel screen; bidirectional map expected.
  • LeetCode Discuss (2026-Q1)— Listed in Vercel interview prep notes.

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 characters.

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-direction mapping

Map each s[i] -> t[i]; if it ever conflicts, return false.

Time
O(n)
Space
O(1) for ASCII
// WRONG — only checks one direction.
function isIsomorphic(s, t) {
  const map = {};
  for (let i = 0; i < s.length; i++) {
    if (map[s[i]] && map[s[i]] !== t[i]) return false;
    map[s[i]] = t[i];
  }
  return true;
}
// Fails for s='ab', t='aa' — 'a' and 'b' both map to 'a'.

Tradeoff: Misses the 'no two source characters map to the same target' rule. Vercel will catch this.

2. Bidirectional mapping (optimal)

Two maps: s[i] -> t[i] and t[i] -> s[i]. Both must be consistent.

Time
O(n)
Space
O(1) for ASCII
function isIsomorphic(s, t) {
  if (s.length !== t.length) return false;
  const sToT = {};
  const tToS = {};
  for (let i = 0; i < s.length; i++) {
    const a = s[i], b = t[i];
    if (sToT[a] && sToT[a] !== b) return false;
    if (tToS[b] && tToS[b] !== a) return false;
    sToT[a] = b;
    tToS[b] = a;
  }
  return true;
}

Tradeoff: Two maps catch both directions of the 1-1 constraint. O(1) extra space because ASCII is bounded.

Vercel-specific tips

Vercel grades the bidirectional check. Bonus signal: noting that 'isomorphic' implies a bijection between the character sets used, and that single-direction mapping is the most common bug. They may follow up by asking 'what if it's a homomorphism instead of isomorphism?' to test definitions.

Common mistakes

  • Only checking s -> t — fails for ('ab', 'aa').
  • Using `if (map[s[i]])` and missing the case where the mapped char is the empty string (won't happen in ASCII but flag it).
  • Initializing the map as `new Map()` and forgetting Map.has() vs key lookup.

Follow-up questions

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

  • Word Pattern (LC 290) — same shape with words instead of characters.
  • Generalize to k-isomorphism (allow up to k mismatches).
  • Find the largest isomorphic substring.

Solve it now

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Output

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FAQ

Why two maps and not one?

Isomorphism requires a bijection. One map only enforces 'each s character maps to one t character.' The second map enforces 'each t character is mapped FROM at most one s character' — the 1-1 part.

Why is space O(1)?

The alphabet is ASCII (256 characters). The maps have at most 256 entries each, independent of n.

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