Word Break Problem in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Word Break DP Table

What is the output of the DP table after running the word break algorithm on the string "leetcode" with dictionary {"leet", "code"}?

DSA C++

bool wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> dict(wordDict.begin(), wordDict.end());
    vector<bool> dp(s.size() + 1, false);
    dp[0] = true;
    for (int i = 1; i <= s.size(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && dict.count(s.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    for (bool val : dp) cout << (val ? 1 : 0) << " ";
    return dp[s.size()];
}

// Called with s = "leetcode", wordDict = {"leet", "code"}

A 1 0 0 1 1 0 0 0 1

B 1 0 0 0 1 0 0 0 1

C 1 0 0 0 1 0 0 1 1

D 1 0 0 0 0 0 0 0 1

💡 Hint

Trace the DP array from start to end, marking positions where substrings are found in the dictionary.

The DP array starts with dp[0] = true. 'leet' is found from index 0 to 4, so dp[4] = true. Then 'code' is found from index 4 to 8, so dp[8] = true. The rest remain false.

🧠 Conceptual

intermediate

Understanding Word Break Problem Complexity

What is the time complexity of the classic dynamic programming solution to the Word Break problem for a string of length n and a dictionary of size m?

A O(n^2 * m)

B O(m^2)

C O(n^3)

D O(n * m)

💡 Hint

Consider the nested loops and substring checks involved in the DP solution.

The DP solution uses two nested loops over the string length n, and for each substring, it checks membership in the dictionary which can be O(m) if implemented naively.

🔧 Debug

advanced

Identify the Bug in Word Break Implementation

Given the following code snippet, what error will it cause when run with s = "applepenapple" and wordDict = {"apple", "pen"}? bool wordBreak(string s, vector& wordDict) { unordered_set dict(wordDict.begin(), wordDict.end()); vector dp(s.size(), false); dp[0] = true; for (int i = 1; i <= s.size(); i++) { for (int j = 0; j < i; j++) { if (dp[j] && dict.count(s.substr(j, i - j))) { dp[i] = true; break; } } } return dp[s.size()]; }

A Logical error: always returns false

B Compilation error due to missing header

C No error, returns true

D Index out of range error

💡 Hint

Check the size of the dp vector and the indices accessed.

The dp vector is initialized with size s.size(), but the code accesses dp[s.size()] which is out of range causing runtime error.

🚀 Application

advanced

Resulting Segmentation Count

How many different ways can the string "catsanddog" be segmented into dictionary words from {"cat", "cats", "and", "sand", "dog"}?

A 2

B 3

C 1

D 4

💡 Hint

Try to find all valid segmentations by splitting the string at dictionary words.

The two segmentations are "cats and dog" and "cat sand dog".

❓ Predict Output

expert

Output of Word Break with Overlapping Dictionary Words

What is the output of the following code snippet? string s = "pineapplepenapple"; vector wordDict = {"apple", "pen", "applepen", "pine", "pineapple"}; bool wordBreak(string s, vector& wordDict) { unordered_set dict(wordDict.begin(), wordDict.end()); vector dp(s.size() + 1, false); dp[0] = true; for (int i = 1; i <= s.size(); i++) { for (int j = 0; j < i; j++) { if (dp[j] && dict.count(s.substr(j, i - j))) { dp[i] = true; break; } } } return dp[s.size()]; } cout << (wordBreak(s, wordDict) ? "true" : "false") << endl;

A Compilation error

B false

C true

D Runtime error

💡 Hint

Consider that multiple dictionary words can overlap and still allow full segmentation.

The string can be segmented as "pine apple pen apple", "pineapple pen apple", or "pine applepen apple", so the function returns true.

Practice

1. What does the Word Break Problem ask you to determine about a given string s and a dictionary of words wordDict?

easy

A. If s is exactly equal to one of the words in wordDict

B. If s contains any substring from wordDict

C. If s can be segmented into a space-separated sequence of one or more dictionary words

D. If s is a palindrome formed by dictionary words

Solution

1. Step 1: Understand the problem statement

   The Word Break Problem asks if the entire string s can be split into parts where each part is a word from the dictionary wordDict.

2. Step 2: Differentiate from similar concepts

   It is not enough for s to contain a dictionary word or be equal to one; the whole string must be fully segmented into dictionary words.

3. Final Answer:

   If s can be segmented into a sequence of dictionary words -> Option C

4. Quick Check:

   Word Break = Full segmentation check [OK]

Hint: Check if full string splits into dictionary words [OK]

Common Mistakes:

• Thinking partial substring match is enough
• Confusing exact match with segmentation
• Ignoring the order of words in segmentation

2. Which of the following is the correct C++ syntax to declare a function named wordBreak that takes a string s and a vector<string> named wordDict and returns a bool?

easy

A. bool wordBreak(string s, vector<string> &wordDict);

B. bool wordBreak(string s, vector<string>& wordDict);

C. bool wordBreak(string s, vector<string> wordDict);

D. bool wordBreak(string s, vector<string> wordDict&);

Solution

1. Step 1: Understand parameter passing in C++

   Passing vector<string> by reference requires the ampersand & after the type, with optional space before the variable name.

2. Step 2: Check syntax correctness

   bool wordBreak(string s, vector<string> &wordDict); correctly places the ampersand before the parameter name with a space: vector<string> &wordDict. bool wordBreak(string s, vector<string>& wordDict); is also valid but less common spacing. bool wordBreak(string s, vector<string> wordDict&); is invalid syntax.

3. Final Answer:

   bool wordBreak(string s, vector<string> &wordDict); -> Option A

4. Quick Check:

   Reference parameter syntax = bool wordBreak(string s, vector<string> &wordDict); [OK]

Hint: Ampersand & goes between type and variable name [OK]

Common Mistakes:

• Placing & after variable name incorrectly
• Missing semicolon at function declaration
• Using invalid syntax like wordDict&

3. What is the output of the following C++ code snippet?

bool wordBreak(string s, vector<string> &wordDict) {
    unordered_set<string> dict(wordDict.begin(), wordDict.end());
    vector<bool> dp(s.size() + 1, false);
    dp[0] = true;
    for (int i = 1; i <= s.size(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && dict.count(s.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.size()];
}

int main() {
    string s = "leetcode";
    vector<string> wordDict = {"leet", "code"};
    cout << (wordBreak(s, wordDict) ? "true" : "false") << endl;
    return 0;
}

medium

A. true

B. false

C. Compilation error

D. Runtime error

Solution

1. Step 1: Understand the dynamic programming logic

   The code uses a boolean vector dp where dp[i] is true if substring s[0..i-1] can be segmented into dictionary words. It initializes dp[0] as true (empty string).

2. Step 2: Trace the example input "leetcode" with dictionary ["leet", "code"]

   At i=4, substring "leet" is in dictionary, so dp[4] = true. At i=8, substring "code" is in dictionary and dp[4] is true, so dp[8] = true. Finally, dp[s.size()] = true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   DP segmentation confirms full string = true [OK]

Hint: Check dp array for full string index true [OK]

Common Mistakes:

• Confusing substring indices
• Forgetting to break inner loop after true
• Misunderstanding dp initialization

4. Identify the bug in the following C++ code for the Word Break Problem:

bool wordBreak(string s, vector<string> &wordDict) {
    unordered_set<string> dict(wordDict.begin(), wordDict.end());
    vector<bool> dp(s.size(), false);
    dp[0] = true;
    for (int i = 1; i <= s.size(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && dict.count(s.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.size()];
}

medium

A. dp[0] should be initialized to false

B. dp vector size should be s.size() + 1, not s.size()

C. The substring indices in s.substr are incorrect

D. unordered_set should be a vector instead

Solution

1. Step 1: Check dp vector size and indexing

   The dp vector is sized to s.size(), but the code accesses dp[s.size()] which is out of bounds. It should be sized s.size() + 1 to include the base case for empty substring.

2. Step 2: Verify other parts

   unordered_set usage and substring indices are correct. dp[0] must be true to represent empty string segmentation.

3. Final Answer:

   dp vector size should be s.size() + 1, not s.size() -> Option B

4. Quick Check:

   dp size off by one causes runtime error [OK]

Hint: dp size = string length + 1 for base case [OK]

Common Mistakes:

• Using dp size equal to string length only
• Misinitializing dp[0] to false
• Confusing substring parameters

5. Given the string s = "applepenapple" and dictionary wordDict = ["apple", "pen"], which of the following modifications to the Word Break algorithm allows it to also return the actual segmented sentence "apple pen apple" instead of just true or false?

hard

A. Use a stack to store indices and print words in reverse order after DP

B. Return the count of possible segmentations instead of a boolean

C. Modify the dictionary to include spaces between words

D. Use a vector of strings to store sentences at each index and build them during DP

Solution

1. Step 1: Understand the limitation of boolean DP

   The standard Word Break DP returns only true/false indicating if segmentation is possible, but does not store the actual segmented sentences.

2. Step 2: Modify DP to store sentences

   By using a vector of strings at each dp index, we can store all possible sentences formed up to that index. When a valid word is found, append it to sentences from previous indices.

3. Step 3: Final answer

   This approach allows reconstructing the segmented sentence like "apple pen apple".

4. Final Answer:

   Use a vector of strings to store sentences at each index and build them during DP -> Option D

5. Quick Check:

   Store sentences in DP to reconstruct output [OK]

Hint: Store sentences in DP array to reconstruct segmentation [OK]

Common Mistakes:

• Trying to modify dictionary words
• Returning count instead of sentences
• Using stack without DP sentence storage