Question

Formatted question description: https://leetcode.ca/all/290.html

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.

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

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.

Algorithm

If the one-to-one mapping cannot be established, return false directly.

It also returns false when the mapping values of the two HashMaps are not the same.

Code

import java.util.HashMap;
import java.util.Map;

public class Word_Pattern {


    public class Solution {
        public boolean wordPattern(String pattern, String str) {
            if (pattern == null || pattern.length() == 0 || str == null || str.length() == 0) {
                return false;
            }

            String[] tokens = str.split(" ");

            if (pattern.length() != tokens.length) {
                return false;
            }

            Map<String, Character> inverseMap = new HashMap<>();
            Map<Character, String> map = new HashMap<>();

            for (int i = 0; i < pattern.length(); i++) {
                char c = pattern.charAt(i);
                String token = tokens[i];

                // Check the one-one mapping
                if (!map.containsKey(c) && !inverseMap.containsKey(token)) {
                    map.put(c, token);
                    inverseMap.put(token, c);
                } else if (map.containsKey(c) && inverseMap.containsKey(token)) {
                    String tokenToCheck = map.get(c);
                    char charToCheck = inverseMap.get(token);

                    if (!tokenToCheck.equals(token) || c != charToCheck) {
                        return false;
                    }
                } else { // @note: this is for different length of pattern and str
                    return false;
                }
            }

            return true;
        }
    }
}

############

class Solution {
    public boolean wordPattern(String pattern, String s) {
        String[] ss = s.split(" ");
        int n = pattern.length();
        if (n != ss.length) {
            return false;
        }
        Map<Character, String> c2str = new HashMap<>();
        Map<String, Character> str2c = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            char k = pattern.charAt(i);
            String v = ss[i];
            if (c2str.containsKey(k) && !Objects.equals(c2str.get(k), v)) {
                return false;
            }
            if (str2c.containsKey(v) && !Objects.equals(str2c.get(v), k)) {
                return false;
            }
            c2str.put(k, v);
            str2c.put(v, k);
        }
        return true;
    }
}

// OJ: https://leetcode.com/problems/word-pattern/
// Time: O(M + N)
// Space: O(M + N)
class Solution {
public:
    bool wordPattern(string pattern, string str) {
        unordered_map<char, string> m;
        unordered_map<string, char> r;
        istringstream ss(str);
        string word;
        for (char c : pattern) {
            if (!(ss >> word)) return false;
            if ((m.count(c) && m[c] != word)
                || r.count(word) && r[word] != c) return false;
            m[c] = word;
            r[word] = c;
        }
        return ss.eof();
    }
};

'''
>>> p = "abba"
>>> s = "dog cat cat dog".split()
>>> s
['dog', 'cat', 'cat', 'dog']
>>> zip(p,s)
[('a', 'dog'), ('b', 'cat'), ('b', 'cat'), ('a', 'dog')]
>>> set(zip(p,s))
set([('b', 'cat'), ('a', 'dog')])
>>>
>>>
>>> s = "dog dog dog dog".split() # then false for: len(set(pattern)) == len(set(str))
>>> zip(p,s)
[('a', 'dog'), ('b', 'dog'), ('b', 'dog'), ('a', 'dog')]
>>> set(zip(p,s))
set([('a', 'dog'), ('b', 'dog')])
'''

class Solution(object):
    def wordPattern(self, pattern, s):
        """
        :type pattern: str
        :type s: str
        :rtype: bool
        """
        s = s.split()
        a = zip(pattern, s)
        return len(pattern) == len(s) and len(set(a)) == len(set(pattern)) == len(set(s))

############

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        s = s.split(' ')
        n = len(pattern)
        if n != len(s):
            return False
        c2str, str2c = defaultdict(), defaultdict()
        for i in range(n):
            k, v = pattern[i], s[i]
            if k in c2str and c2str[k] != v:
                return False
            if v in str2c and str2c[v] != k:
                return False
            c2str[k], str2c[v] = v, k
        return True

func wordPattern(pattern string, s string) bool {
	ss := strings.Split(s, " ")
	n := len(pattern)
	if n != len(ss) {
		return false
	}
	c2str := make(map[byte]string)
	str2c := make(map[string]byte)
	for i := 0; i < n; i++ {
		k, v := pattern[i], ss[i]
		if c2str[k] != "" && c2str[k] != v {
			return false
		}
		if str2c[v] > 0 && str2c[v] != k {
			return false
		}
		c2str[k], str2c[v] = v, k
	}
	return true
}

function wordPattern(pattern: string, s: string): boolean {
    let n = pattern.length;
    let values = s.split(' ');
    if (n != values.length) return false;
    let table = new Array(128);
    for (let i = 0; i < n; i++) {
        let k = pattern.charCodeAt(i),
            v = values[i];
        if (!table[k]) {
            if (table.includes(v)) return false;
            table[k] = v;
        } else {
            if (table[k] != v) return false;
        }
    }
    return true;
}

use std::collections::HashMap;

impl Solution {
    pub fn word_pattern(pattern: String, s: String) -> bool {
        let cs1: Vec<char> = pattern.chars().collect();
        let cs2: Vec<&str> = s.split_whitespace().collect();
        let n = cs1.len();
        if n != cs2.len() {
            return false;
        }
        let mut map1 = HashMap::new();
        let mut map2 = HashMap::new();
        for i in 0..n {
            let c = cs1[i];
            let s = cs2[i];
            if !map1.contains_key(&c) {
                map1.insert(c, i);
            }
            if !map2.contains_key(&s) {
                map2.insert(s, i);
            }
            if map1.get(&c) != map2.get(&s) {
                return false
            }
        }
        true
    }
}

public class Solution {
    public bool WordPattern(string pattern, string s) {
        var ws = s.Split(' ');
        if (pattern.Length != ws.Length) {
            return false;
        }
        var d1 = new Dictionary<char, string>();
        var d2 = new Dictionary<string, char>();
        for (int i = 0; i < ws.Length; ++i) {
            var a = pattern[i];
            var b = ws[i];
            if (d1.ContainsKey(a) && d1[a] != b) {
                return false;
            }
            if (d2.ContainsKey(b) && d2[b] != a) {
                return false;
            }
            d1[a] = b;
            d2[b] = a;
        }
        return true;
    }
}

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