290. Word Pattern

EasyHash TableString
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Problem Description

The problem presents the task of determining if a given string s follows the same pattern as given by a string pattern. Each letter in pattern corresponds to a non-empty word in s, and the relationship between the letters and the words must be a bijection. This means every character in the pattern should map to a uniquely associated word in s and vice versa, with no two letters mapping to the same word and no two words mapping to the same letter.


The key to solving this problem is to maintain a mapping between the characters of the pattern and the words in s. We can achieve this by using two hash tables: one table (d1) to map characters to words and another (d2) to map words to characters.

We start by splitting the string s into words. The number of words in s should match the number of characters in the pattern; if not, the pattern cannot be followed, and we can return false immediately. After checking this length criterion, we proceed to iterate over the characters and words in parallel.

For every character-word pair, we check:

  1. If the current character is already mapped to a different word in d1, the pattern is broken.
  2. Conversely, if the current word is already mapped to a different character in d2, the pattern is also broken.

If none of the above cases are true, we record the mapping in both directions (d1 and d2). This enforces the bijection principle. If we can complete this process without conflicts, then the string s successfully follows the pattern defined by pattern, and we return true.

Solution Approach

The solution follows a straightforward approach using hash tables. Here's how it works, step by step:

  1. We begin by splitting the input string s using the split() method, which returns a list of words. This list is stored in a variable ws.

  2. We immediately check if the number of words in ws is equal to the length of the pattern. If there's a mismatch, we return false as it's not possible for s to follow the pattern if the number of elements doesn't match.

  3. We declare two dictionaries, d1 and d2. d1 will keep track of the letter-to-word mappings, and d2 will keep track of word-to-letter mappings. They help us verify that each letter in pattern maps to exactly one word in s and each word in s maps to exactly one letter in pattern.

  4. We use the built-in zip function to iterate over the characters in pattern and the words in ws simultaneously. For each character a and word b:

    • We check if a is already a key in d1. If a is already mapped to a word, and that word is not b, we have a conflict, meaning s does not follow the pattern, and we return false.

    • Similarly, we check if b is already a key in d2. If b is already mapped to a character, and that character is not a, we again have a conflict, and we return false.

  5. If neither of the above conflicts arises, we map character a to word b in d1 and word b to character a in d2.

  6. If we can iterate over all character-word pairs without encountering any conflicts, the function returns true, indicating that the string s does follow the given pattern.

This approach cleverly uses the properties of hash tables: constant time complexity for insertions and lookups (on average). By maintaining two separate mappings and checking for existing mappings at each step, we ensure a bijection between the characters of pattern and the words in s. This algorithm has a time complexity of O(n), where n is the number of characters in pattern or the number of words in s, assuming that the hash operations are constant time.

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Example Walkthrough

To illustrate the solution approach, let's consider an example where the pattern string is "abba" and the string s is "dog cat cat dog".

  1. We begin by splitting the string s into words, which gives us ws = ["dog", "cat", "cat", "dog"].

  2. Next, we check whether the length of ws matches the length of pattern. Here, they both have a length of 4, so we can proceed with the mapping process.

  3. We initialize two dictionaries: d1 = {} and d2 = {}.

  4. Start iterating through the pairs generated by zip("abba", ["dog", "cat", "cat", "dog"]), which gives us:

    • (a, dog)
    • (b, cat)
    • (b, cat)
    • (a, dog)
  5. For the first pair (a, dog), a is not a key in d1 and dog is not a key in d2, so we add them to our dictionaries:

    • d1 = {'a': 'dog'}
    • d2 = {'dog': 'a'}
  6. For the second pair (b, cat), b is not a key in d1 and cat is not a key in d2, so we likewise add them:

    • d1 = {'a': 'dog', 'b': 'cat'}
    • d2 = {'dog': 'a', 'cat': 'b'}
  7. For the third pair (b, cat), b is already in d1 and it maps to cat, and cat is in d2 and it maps to b, so no conflict occurs.

  8. Finally, for the fourth pair (a, dog), a is already in d1 and it maps to dog, and dog is in d2 and it maps to a, so again no conflict occurs.

Since we have iterated through all character-word pairs without encountering any conflicts, we can conclude that the string s does indeed follow the given pattern. Therefore, our function should return true for this example.

Solution Implementation

1class Solution:
2    def wordPattern(self, pattern: str, str_sequence: str) -> bool:
3        # Split the input string on whitespace to get individual words
4        words = str_sequence.split()
6        # If the pattern length and word count are different, they don't match
7        if len(pattern) != len(words):
8            return False
10        # Initialize dictionaries to store character-to-word map and word-to-character map
11        char_to_word_map = {}
12        word_to_char_map = {}
14        # Iterate over the pattern and the corresponding words together
15        for char, word in zip(pattern, words):
16            # If the character is already mapped to a different word,
17            # or the word is already mapped to a different character, return False
18            if (char in char_to_word_map and char_to_word_map[char] != word) or \
19               (word in word_to_char_map and word_to_char_map[word] != char):
20                return False
22            # Add the mappings to both dictionaries
23            char_to_word_map[char] = word
24            word_to_char_map[word] = char
26        # If no mismatches are found, pattern and words match - return True
27        return True
1class Solution {
2    public boolean wordPattern(String pattern, String s) {
3        // Split the s string into individual words
4        String[] words = s.split(" ");
6        // If the number of characters in the pattern does not match the number of words, return false
7        if (pattern.length() != words.length) {
8            return false;
9        }
11        // Initialize two dictionaries to track the mappings from characters to words and vice versa
12        Map<Character, String> charToWordMap = new HashMap<>();
13        Map<String, Character> wordToCharMap = new HashMap<>();
15        // Iterate over the pattern
16        for (int i = 0; i < words.length; ++i) {
17            char currentChar = pattern.charAt(i);
18            String currentWord = words[i];
20            // If the current mapping from char to word or word to char does not exist or is inconsistent, return false
21            if (!charToWordMap.getOrDefault(currentChar, currentWord).equals(currentWord) || wordToCharMap.getOrDefault(currentWord, currentChar) != currentChar) {
22                return false;
23            }
25            // Update the mappings
26            charToWordMap.put(currentChar, currentWord);
27            wordToCharMap.put(currentWord, currentChar);
28        }
30        // If no inconsistencies are found, return true
31        return true;
32    }
1#include <sstream>
2#include <vector>
3#include <string>
4#include <unordered_map>
6class Solution {
8    // Determines if a pattern matches the words in a string
9    bool wordPattern(string pattern, string str) {
10        // Utilize istringstream to split the string into words
11        istringstream strStream(str);
12        vector<string> words;
13        string word;
15        // Splitting the string by whitespaces
16        while (strStream >> word) {
17            words.push_back(word);
18        }
20        // If the number of pattern characters and words do not match, return false
21        if (pattern.size() != words.size()) {
22            return false;
23        }
25        // Create mappings to keep track of the pattern to word relationships
26        unordered_map<char, string> patternToWord;
27        unordered_map<string, char> wordToPattern;
29        // Iterate through the pattern and corresponding words
30        for (int i = 0; i < words.size(); ++i) {
31            char patternChar = pattern[i];
32            string currentWord = words[i];
34            // Check if the current pattern character has already been mapped to a different word
35            // or the current word has been mapped to a different pattern character
36            if ((patternToWord.count(patternChar) && patternToWord[patternChar] != currentWord) ||
37                (wordToPattern.count(currentWord) && wordToPattern[currentWord] != patternChar)) {
38                return false;
39            }
41            // Map the current pattern character to the current word and vice versa
42            patternToWord[patternChar] = currentWord;
43            wordToPattern[currentWord] = patternChar;
44        }
46        // If all pattern characters and words match up, return true
47        return true;
48    }
1// Checks if a string pattern matches a given string sequence.
2// Each letter in the pattern corresponds to a word in the s string.
3function wordPattern(pattern: string, s: string): boolean {
4    // Split the string s into an array of words.
5    const words = s.split(' ');
7    // If the number of elements in the pattern does not match the number of words, return false.
8    if (pattern.length !== words.length) {
9        return false;
10    }
12    // Initialize two maps to store the character-to-word and word-to-character correspondences.
13    const charToWordMap = new Map<string, string>();
14    const wordToCharMap = new Map<string, string>();
16    // Iterate over the pattern.
17    for (let i = 0; i < pattern.length; ++i) {
18        const char = pattern[i]; // Current character from the pattern.
19        const word = words[i]; // Current word from the string.
21        // Check if the current character is already associated with a different word.
22        if (charToWordMap.has(char) && charToWordMap.get(char) !== word) {
23            return false; // Mismatch found, return false.
24        }
26        // Check if the current word is already associated with a different character.
27        if (wordToCharMap.has(word) && wordToCharMap.get(word) !== char) {
28            return false; // Mismatch found, return false.
29        }
31        // Add the current character-to-word and word-to-character association to the maps.
32        charToWordMap.set(char, word);
33        wordToCharMap.set(word, char);
34    }
36    // If no mismatch was found, return true.
37    return true;

Time and Space Complexity

The function wordPattern checks if a string follows a specific pattern. The time complexity and space complexity analysis is as follows:

Time Complexity

The time complexity of the code is O(N) where N is the length of the longer of the two inputs: the pattern and the s.split() list. s.split() operation itself is O(n) where n is the length of the string s, as it must traverse the string once and create a list of words.

The zip(pattern, ws) operation will iterate over the pairs of characters in pattern and words in ws, and the number of iterations will be the lesser of the two lengths. However, since we've already ensured both lengths are equal, it results in min(len(pattern), len(ws)) operations, which in this case is len(pattern) or equivalently len(ws).

Inside the loop, the operations involve checking membership and equality in two dictionaries, d1 and d2. Dictionary membership and assignment are average case O(1) operations due to hash table properties.

Combining these operations results in the total time complexity being O(N), where N = max(m, n) (the longer of the pattern's length m and the split string list ws length n). But since the function only works when m == n, you can simplify the statement to just O(n) where n is the length of pattern or ws.

Space Complexity

The space complexity of the function is also O(N) where N = m + n, the size of the input pattern plus the size of the list ws, which comes from the split() of string s.

The two dictionaries d1 and d2 will store at most m keys (unique characters in the pattern) and n keys (unique words in s).

Provided that both pattern and s.split() are of the same length due to the early return conditional, the space complexity simplifies to O(n), where n is the length of pattern or ws.

Learn more about how to find time and space complexity quickly using problem constraints.

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