Word Pattern

LeetCode 290 | Difficulty: Easy

Easy

Problem Description

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. Specifically:

- Each letter in `pattern` maps to **exactly** one unique word in `s`.

- Each unique word in `s` maps to **exactly** one letter in `pattern`.

- No two letters map to the same word, and no two words map to the same letter.

Example 1:

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

Output: true

Explanation:

The bijection can be established as:

- `'a'` maps to `"dog"`.

- `'b'` maps to `"cat"`.

Example 2:

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

Output: false

Example 3:

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

Output: false

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**.

Topics: Hash Table, String

---

Approach

Hash Map

Use a hash map for O(1) average lookups. Store seen values, frequencies, or indices. The key question: what should I store as key, and what as value?

When to use

Need fast lookups, counting frequencies, finding complements/pairs.

String Processing

Consider character frequency counts, two-pointer approaches, or building strings efficiently. For pattern matching, think about KMP or rolling hash. For palindromes, expand from center or use DP.

When to use

Anagram detection, palindrome checking, string transformation, pattern matching.

---

Solutions

Solution 1: C# (Best: 108 ms)

Metric	Value
Runtime	108 ms
Memory	36.5 MB
Date	2022-02-08

Solution

public class Solution {
    public bool WordPattern(string pattern, string s) {
        Dictionary<string, int> d = new Dictionary<string, int>();
    string[] words = new string[26];
    string[] arr = s.Split(new char[] { ' ' }).ToArray();
    if (pattern.Length != arr.Length) return false;

    for (int i = 0; i < pattern.Length; i++)
    {
        int index = pattern[i] - 'a';
        if (string.IsNullOrEmpty(words[index]))
        {
            if(d.ContainsKey(arr[i])) return false;
            words[index] = arr[i];
            d.Add(arr[i], index);
        }
        else if (arr[i] != words[index])
            return false;
    }

    return true;
    }
}

---

Complexity Analysis

Approach	Time	Space
Hash Map	$O(n)$	$O(n)$

---

Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Hash map gives O(1) lookup — think about what to use as key vs value.