Word Break Problem Statement

Given a string and a dictionary of words, check whether the given string can be broken down into a space-separated sequence of one or more dictionary words.

Example One

{
"s": "helloworldhello",
"words_dictionary": ["world", "hello", "faang"]
}

Output:

1

helloworldhello can be broken down as hello world hello.

Example Two

{
"s": "interviewkickstart",
"words_dictionary": ["interview", "preparation"]
}

Output:

0

Notes

The same word in the dictionary may be used multiple times in the breakdown process.

Constraints:

• 1 <= length of given string <= 10³
• 1 <= number of words in the dictionary <= 10³
• 1 <= length of each word in the dictionary <= 20
• Each string consists of lowercase English alphabets only.
• All words in the dictionary are unique.

Code For Word Break Solution: Bottom Up Dp

/*
Asymptotic complexity in terms of length of string `n`, length of dictionary `m` and maximum length of any word in the
dictionary `w`:
* Time: O(n^2 * w).
* Auxiliary space: O(n + m * w).
* Total space: O(n + m * w).
*/
bool word_break(string &s, vector<string> &words_dictionary) {
    unordered_set<string> words_set(words_dictionary.begin(), words_dictionary.end());
    int n = s.length();
    // dp[i] stores a boolean value, indicating whether it is possible to break a string starting from index i to n - 1.
    vector<bool> dp(n + 1, false);
    dp[n] = true;

    for (int i = n - 1; i >= 0; i--) {
        string word;
        for (int j = i; j < n; j++) {
            word += s[j];
            // As the maximum word length in the dictionary is 20.
            if (word.length() > 20) {
                break;
            }
            if (dp[j + 1] and words_set.find(word) != words_set.end()) {
                dp[i] = true;
                break;
            }
        }
    }

    return dp[0];
}

We hope that these solutions to word break problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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