Word Break Problem in DSA C - Time & Space Complexity

Time Complexity: Word Break Problem

O(n^2 * m * k)

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(char* s, char** wordDict, int wordDictSize) {
    int n = strlen(s);
    bool dp[n+1];
    dp[0] = true;
    for (int i = 1; i <= n; i++) {
        dp[i] = false;
        for (int j = 0; j < i; j++) {
            if (dp[j] && isInDict(s + j, i - j, wordDict, wordDictSize)) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[n];
}

// Helper function checks if substring s of length len is in wordDict
bool isInDict(char* s, int len, char** wordDict, int wordDictSize) {
    for (int k = 0; k < wordDictSize; k++) {
        if (strlen(wordDict[k]) == len && strncmp(s, wordDict[k], len) == 0) {
            return true;
        }
    }
    return false;
}
    

This code checks if the input string can be split into words from the dictionary using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops where for each position in the string, we check all previous positions.
• How many times: Outer loop runs n times; inner loop runs up to n times for each outer iteration.
• Additional work: For each substring checked, we scan the dictionary to find a match.

How Execution Grows With Input

As the string length grows, the number of checks grows roughly with the square of the length times the dictionary size.

Input Size (n)	Approx. Operations
10	About 10 * 10 * wordDictSize checks
100	About 100 * 100 * wordDictSize checks
1000	About 1000 * 1000 * wordDictSize checks

Pattern observation: Operations grow roughly with the square of the input length multiplied by dictionary size.

Final Time Complexity

Time Complexity: O(n^2 * m * k)

This means the time grows roughly with the square of the string length (n), times the number of words in the dictionary (m), times the average length of dictionary words (k).

Common Mistake

[X] Wrong: "The solution runs in linear time because it just loops through the string once."

[OK] Correct: The code uses nested loops and checks many substrings against the dictionary, so the work grows much faster than just one pass.

Interview Connect

Understanding this time complexity helps you explain how your solution scales and shows you can analyze nested loops and dictionary lookups clearly.

Self-Check

"What if we used a hash set for the dictionary to check words in constant time? 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 contains any substring from wordDict

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

C. If s is exactly equal to any word in wordDict

D. If s is a palindrome

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 in the dictionary.

2. Step 2: Compare options with the problem

   If s can be segmented into a space-separated sequence of one or more dictionary words correctly describes this segmentation into dictionary words. Other options describe different checks.

3. Final Answer:

   If s can be segmented into a space-separated sequence of one or more dictionary words -> Option B

4. Quick Check:

   Word Break Problem = segmentation check [OK]

Hint: Remember: Word Break means full segmentation into dictionary words [OK]

Common Mistakes:

• Thinking it checks for substring presence only
• Confusing exact match with segmentation
• Assuming palindrome check

2. Which of the following C code snippets correctly declares a function prototype for solving the Word Break Problem?

easy

A. int wordBreak(char s[], char *wordDict[], int wordDictSize);

B. void wordBreak(char s[], char wordDict[], int size);

C. int wordBreak(char *s, char *wordDict, int wordDictSize);

D. int wordBreak(char **s, char **wordDict, int wordDictSize);

Solution

1. Step 1: Understand parameter types for the Word Break function

   The string s is a character array, so char s[] or char *s works. The dictionary is an array of strings, so char *wordDict[] is correct. The size is an integer.

2. Step 2: Check each option

   int wordBreak(char s[], char *wordDict[], int wordDictSize); matches the correct signature: int wordBreak(char s[], char *wordDict[], int wordDictSize); int wordBreak(char *s, char *wordDict, int wordDictSize); uses char *wordDict which is incorrect for array of strings. void wordBreak(char s[], char wordDict[], int size); uses char wordDict[] which is incorrect for array of strings. int wordBreak(char **s, char **wordDict, int wordDictSize); uses char **s which is incorrect for a single string.

3. Final Answer:

   int wordBreak(char s[], char *wordDict[], int wordDictSize); -> Option A

4. Quick Check:

   Function prototype with string and array of strings = int wordBreak(char s[], char *wordDict[], int wordDictSize); [OK]

Hint: Dictionary is array of strings: use char *wordDict[] [OK]

Common Mistakes:

• Using char wordDict[] instead of array of strings
• Using char **s for input string
• Wrong return type or missing parameters

3. What will be the output of the following C code snippet for the Word Break Problem?

char *dict[] = {"leet", "code"};
int result = wordBreak("leetcode", dict, 2);
printf("%d", result);

Assuming wordBreak returns 1 if segmentation is possible, else 0.

medium

A. 1

B. 0

C. Compilation error

D. Runtime error

Solution

1. Step 1: Analyze the input string and dictionary

   The string is "leetcode" and dictionary contains "leet" and "code".

2. Step 2: Check if string can be segmented

   "leetcode" can be split as "leet" + "code", both in dictionary, so segmentation is possible.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   "leetcode" segmented as "leet" + "code" = 1 [OK]

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

Common Mistakes:

• Assuming partial match is enough
• Confusing return values
• Ignoring dictionary contents

4. Identify the error in the following C code snippet for the Word Break Problem:

int wordBreak(char *s, char **wordDict, int wordDictSize) {
    int len = strlen(s);
    int dp[len+1];
    dp[0] = 1;
    for (int i = 1; i <= len; i++) {
        dp[i] = 0;
        for (int j = 0; j < i; j++) {
            if (dp[j] && strncmp(s+j, wordDict[i], i-j) == 0) {
                dp[i] = 1;
                break;
            }
        }
    }
    return dp[len];
}

medium

A. Missing return statement

B. dp array not initialized properly

C. Using wordDict[i] instead of wordDict[j] in strncmp

D. Loop indices are out of range

Solution

1. Step 1: Understand the nested loop and string comparison

   The inner loop checks substrings from index j to i. It should compare with wordDict[j], not wordDict[i].

2. Step 2: Identify the incorrect array index usage

   Using wordDict[i] is wrong because i goes up to length of s, which may exceed wordDictSize and is unrelated to dictionary indexing.

3. Final Answer:

   Using wordDict[i] instead of wordDict[j] in strncmp -> Option C

4. Quick Check:

   Index in dictionary should match substring start index [OK]

Hint: Check indices used for dictionary access in loops [OK]

Common Mistakes:

• Mixing loop variables for dictionary access
• Not initializing dp array correctly
• Ignoring return statement

5. Given the string s = "applepenapple" and dictionary wordDict = ["apple", "pen"], which approach correctly solves the Word Break Problem using dynamic programming in C?

hard

A. Use recursion without memoization to check all substring splits

B. Use a stack to store all dictionary words and pop when matched

C. Sort the dictionary and check if s starts with any word

D. Use a boolean dp array where dp[i] is true if s[0..i-1] can be segmented, then check all substrings ending at i

Solution

1. Step 1: Understand dynamic programming approach

   We create a boolean dp array where dp[i] is true if substring s[0..i-1] can be segmented into dictionary words.

2. Step 2: Check all substrings ending at each position

   For each position i, check all j < i where dp[j] is true and substring s[j..i-1] is in dictionary. If yes, set dp[i] true.

3. Step 3: Return dp[length of s]

   If dp at the end is true, the string can be segmented.

4. Final Answer:

   Use a boolean dp array where dp[i] is true if s[0..i-1] can be segmented, then check all substrings ending at i -> Option D

5. Quick Check:

   DP boolean array with substring checks = Use a boolean dp array where dp[i] is true if s[0..i-1] can be segmented, then check all substrings ending at i [OK]

Hint: Use dp array to track segmentation up to each index [OK]

Common Mistakes:

• Using recursion without memo causing timeout
• Sorting dictionary does not help segmentation
• Stack approach is incorrect for this problem