Javascript Coding Test Course, Dictionary Lookup

Solution

To solve this problem, the following steps are necessary:

1. Split the string s by spaces to create a list of words.
2. Check if all words in the list are included in the dictionary.
3. If all words are included, return true; if any word is not included, return
   false.

Implementation

Based on this logic, let’s write the JavaScript code:

function isAllWordsInDictionary(s, dictionary) {
    const words = s.split(" "); // Split words by spaces
    const dictionarySet = new Set(dictionary); // Convert to Set for optimized search

    for (const word of words) {
        if (!dictionarySet.has(word)) { // Check if each word is in the dictionary
            return false; // Return false if any word is not found
        }
    }
    return true; // Return true if all words are found
}

// Example tests
console.log(isAllWordsInDictionary("apple banana", ["apple", "sweet potato", "banana"])); // true
console.log(isAllWordsInDictionary("orange grape", ["apple", "banana"])); // false
    

Code Explanation

In the code above, we perform the following steps:

• s.split(" "): Splits the given string s by spaces to create a list of words.
• new Set(dictionary): Converts the given dictionary array to a Set to remove duplicates and optimize
  search times to O(1).
• We use a for loop to check if each word exists in the dictionarySet.
• If a word does not exist, false is returned; if all words exist, true is returned.

Time Complexity

The time complexity of this algorithm is O(n + m), where n is the number of words in the string s and
m is the number of words in the dictionary. The reason for using a Set is to improve the search speed to enhance
overall performance.

Conclusion

This problem allowed us to effectively use strings and arrays to verify whether the given conditions were met.
When solving algorithm problems, it is always easier to approach them by breaking them down into steps.
Such problems can also be beneficially utilized in other scenarios.