Word Pattern

Published at24 Jul 2026

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Given a pattern string like "abba" and a string s of space-separated words like "dog cat cat dog", determine if s follows the same pattern. That means there is a bijection between each character in the pattern and each word in s. A letter maps to exactly one word, and each word maps to exactly one letter.

This is a classic hash map problem requiring bidirectional mapping. A single hash map is not enough because it only enforces one direction. For example, pattern "abba" with "dog dog dog dog" would be incorrectly accepted with a one-way map (a→dog, b→dog) because both letters map to the same word. A second hash map tracking word→letter catches this violation.

The algorithm splits the words from s and compares the length of the pattern with the word count. Then it iterates through both, building the two maps simultaneously. If at any point a mapping contradicts an existing one, return false. This runs in O(n) time where n is the length of the pattern (or number of words).

Edge cases include empty pattern or empty string, a pattern longer than the word list, and patterns where multiple letters map to the same word.

Example Input & Output

Example 1

Input

"abba", "dog cat cat dog"

Output

true

Explanation

Bijection holds: a→dog, b→cat.

Example 2

Input

"aaaa", "dog cat cat dog"

Output

false

Explanation

Each a maps to different words.

Example 3

Input

"abba", "dog cat cat fish"

Output

false

Explanation

Last word breaks mapping.

Example 4

Input

"abba", "dog dog dog dog"

Output

false

Explanation

b maps to dog but a already maps to dog.

Example 5

Input

"", ""

Output

true

Explanation

Empty pattern and empty string match.

Algorithm Flow

Recommendation Algorithm Flow for Word Pattern

Solution Approach

Split s into words. If lengths differ, return false. Use two hash maps: p2w (pattern char to word) and w2p (word to pattern char). Iterate together, check consistency.

function solution(pattern, s) {
  var words = s.split(' ');
  if (pattern.length !== words.length) return false;
  var p2w = {}, w2p = {};
  for (var i = 0; i < pattern.length; i++) {
    var ch = pattern[i], w = words[i];
    if (p2w[ch] && p2w[ch] !== w) return false;
    if (w2p[w] && w2p[w] !== ch) return false;
    p2w[ch] = w;
    w2p[w] = ch;
  }
  return true;
}

Time O(n), Space O(n).

Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

java

import java.util.*;
class Solution {
    public boolean solution(String pattern, String s) {
        if (pattern.isEmpty() && s.isEmpty()) return true;
        String[] words = s.split(" ");
        if (pattern.length() != words.length) return false;
        Map<Character,String> p2w = new HashMap<>();
        Map<String,Character> w2p = new HashMap<>();
        for (int i = 0; i < pattern.length(); i++) {
            char ch = pattern.charAt(i);
            String w = words[i];
            if (p2w.containsKey(ch) && !p2w.get(ch).equals(w)) return false;
            if (w2p.containsKey(w) && w2p.get(w) != ch) return false;
            p2w.put(ch, w);
            w2p.put(w, ch);
        }
        return true;
    }
}

javascript

function solution(pattern, s) {
  if (!pattern && !s) return true;
  var words = s.split(' ');
  if (pattern.length !== words.length) return false;
  var p2w = {}, w2p = {};
  for (var i = 0; i < pattern.length; i++) {
    var ch = pattern[i], w = words[i];
    if (p2w[ch] && p2w[ch] !== w) return false;
    if (w2p[w] && w2p[w] !== ch) return false;
    p2w[ch] = w;
    w2p[w] = ch;
  }
  return true;
}

php

<?php
function solution($pattern, $s) {
    if ($pattern === '' && $s === '') return true;
    $words = explode(' ', $s);
    if (strlen($pattern) !== count($words)) return false;
    $p2w = [];
    $w2p = [];
    for ($i = 0; $i < strlen($pattern); $i++) {
        $ch = $pattern[$i];
        $w = $words[$i];
        if (isset($p2w[$ch]) && $p2w[$ch] !== $w) return false;
        if (isset($w2p[$w]) && $w2p[$w] !== $ch) return false;
        $p2w[$ch] = $w;
        $w2p[$w] = $ch;
    }
    return true;
}

python

def solution(pattern, s):
    if not pattern and not s:
        return True
    words = s.split(' ')
    if len(pattern) != len(words):
        return False
    p2w = {}
    w2p = {}
    for ch, w in zip(pattern, words):
        if ch in p2w and p2w[ch] != w:
            return False
        if w in w2p and w2p[w] != ch:
            return False
        p2w[ch] = w
        w2p[w] = ch
    return True

rust

use std::collections::HashMap;
fn solution(pattern: &str, s: &str) -> bool {
    if pattern.is_empty() && s.is_empty() { return true; }
    let words: Vec<&str> = s.split(' ').collect();
    if pattern.len() != words.len() { return false; }
    let mut p2w: HashMap<char, &str> = HashMap::new();
    let mut w2p: HashMap<&str, char> = HashMap::new();
    for (i, ch) in pattern.chars().enumerate() {
        let w = words[i];
        if let Some(&cw) = p2w.get(&ch) { if cw != w { return false; } }
        if let Some(&pc) = w2p.get(w) { if pc != ch { return false; } }
        p2w.insert(ch, w);
        w2p.insert(w, ch);
    }
    true
}

typescript

function solution(pattern: string, s: string): boolean {
    if (!pattern && !s) return true;
    var words = s.split(' ');
    if (pattern.length !== words.length) return false;
    var p2w: any = {}, w2p: any = {};
    for (var i = 0; i < pattern.length; i++) {
        var ch = pattern.charAt(i);
        var w = words[i];
        if (p2w[ch] && p2w[ch] !== w) return false;
        if (w2p[w] && w2p[w] !== ch) return false;
        p2w[ch] = w;
        w2p[w] = ch;
    }
    return true;
}

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Word Pattern

Example Input & Output

Algorithm Flow

Solution Approach

Best AnswersThese are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.

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Best Answers
These are the verified best solutions for this Challenge. You can use them as a reference to improve your solution.