77. Word Pattern

mediumAsked at Salesforce

Determine if a string follows the same pattern as a sequence of words. Salesforce uses this to test bidirectional hashmap invariants in a more complex setting.

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

Source citations

Public interview reports confirming this problem appears in Salesforce loops.

Glassdoor (2026-Q1)— Salesforce uses bijective mapping in their field-mapping wizard.
LeetCode Discuss (2025)— Pairs with LC 205 (Isomorphic Strings).

Problem

Given a pattern and a string s, find if s follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.

Constraints

1 <= pattern.length <= 300
pattern contains only lower-case English letters.
1 <= s.length <= 3000
s contains only lowercase English letters and spaces ' '.
s does not contain any leading or trailing spaces.
All the words in s are separated by a single space.

Examples

Example 1

Input

pattern = "abba", s = "dog cat cat dog"

Output

true

Example 2

Input

pattern = "abba", s = "dog cat cat fish"

Output

false

Example 3

Input

pattern = "aaaa", s = "dog cat cat dog"

Output

false

Approaches

1. Single-direction map (broken)

Map pattern[i] to words[i]; check consistency.

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

// Fails on 'abba', 'dog dog dog dog' — pattern 'a' maps to 'dog' consistently; 'b' maps to 'dog' too, but a and b should map to different words.

Tradeoff: Same trap as LC 205.

2. Bidirectional map

Maintain pattern->word and word->pattern maps; both must be consistent.

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

function wordPattern(pattern, s) {
  const words = s.split(' ');
  if (pattern.length !== words.length) return false;
  const pToW = new Map(), wToP = new Map();
  for (let i = 0; i < pattern.length; i++) {
    const p = pattern[i], w = words[i];
    if (pToW.has(p) && pToW.get(p) !== w) return false;
    if (wToP.has(w) && wToP.get(w) !== p) return false;
    pToW.set(p, w);
    wToP.set(w, p);
  }
  return true;
}

Tradeoff: Bijective check via two maps. The length check upfront prevents off-by-one when pattern and words don't align.

Salesforce-specific tips

Salesforce uses this exact bijective mapping in their field-mapping wizard (each source field must map to exactly one target field and vice versa). They grade specifically on detecting the 'aaaa' → 'dog cat cat dog' case where forward map is fine but reverse fails. Bonus signal: discuss Word Pattern II (LC 291) which extends with backtracking.

Common mistakes

Forgetting to check pattern.length === words.length — gives false positives on length mismatch.
Single-direction map — same trap as LC 205.
Using object keys for words — works in JS but breaks if any word collides with Object prototype (e.g., 'toString').

Follow-up questions

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

Word Pattern II (LC 291) — pattern matches without space delimiters.
Isomorphic Strings (LC 205).
Generalize to multi-character patterns.

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Output

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FAQ

Why two maps?

Because bijectivity is two-way. Forward map enforces 'pattern char → word' consistency. Reverse map enforces 'word → pattern char' consistency.

Why split on space?

The problem says words are space-separated. split(' ') handles single-space delimiters correctly; for multi-space, use split(/\s+/).

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