Word Break Problem in DSA C++ - Time & Space Complexity

Time Complexity: Word Break Problem

O(n^3)

Understanding Time Complexity

We want to understand how the time needed to solve the Word Break Problem changes as the input grows.

Specifically, how does checking if a string can be split into dictionary words scale with string length?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if the string can be segmented into dictionary words using dynamic programming.

Identify Repeating Operations

• Primary operation: Nested loops where the outer loop runs from 1 to n and the inner loop runs from 0 to i-1.
• How many times: The inner loop runs approximately i times for each i, so total checks are about 1 + 2 + ... + n = n(n+1)/2 times.
• Substring checks: Each iteration extracts a substring and checks if it is in the dictionary.

How Execution Grows With Input

As the string length grows, the number of substring checks grows roughly with the square of the length.

Input Size (n)	Approx. Operations
10	About 55 substring checks
100	About 5,050 substring checks
1000	About 500,500 substring checks

Pattern observation: The operations grow roughly like n squared, meaning doubling the input length roughly quadruples the work.

Final Time Complexity

Time Complexity: O(n^3)

This means the time grows roughly with the cube of the string length due to substring extraction and dictionary lookups.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it uses dynamic programming."

[OK] Correct: Although dynamic programming avoids repeated work, the nested loops still check many substrings, causing cubic growth, not linear.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming improves efficiency and why nested loops impact performance.

Self-Check

"What if we used a Trie data structure for the dictionary lookups instead of a hash set? How would the time complexity change?"

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