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### How to Nail your next Technical Interview

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## You may be missing out on a 66.5% salary hike*

### Nick Camilleri

Head of Career Skills Development & Coaching
*Based on past data of successful IK students
Help us know you better!

## How many years of coding experience do you have?

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## FREE course on 'Sorting Algorithms' by Omkar Deshpande (Stanford PhD, Head of Curriculum, IK)

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# 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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

### Try yourself in the Editor

Note: Input and Output will already be taken care of.

# 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.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as: