Word Break Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the input string is empty.

DSA C

if (str[0] == [1]) {
    return 1;
}

Drag options to blanks, or click blank then click option'

A'\0'

B' '

C'a'

D'\n'

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for space ' ' instead of null character.

Using double quotes instead of single quotes for characters.

The string ends with the null character '\0' in C, so checking str[0] == '\0' means the string is empty.

2fill in blank

medium

Complete the code to get the length of the input string.

DSA C

int len = [1](str);

Drag options to blanks, or click blank then click option'

Asizeof

Bstrlen

Clength

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof which returns size of pointer, not string length.

Using non-existent functions like length or count.

In C, the function strlen() returns the length of a string excluding the null character.

3fill in blank

hard

Fix the error in the code that compares a substring with a dictionary word.

DSA C

if (strncmp(str + i, dict[j], [1]) == 0) {
    // substring matches dict word
}

Drag options to blanks, or click blank then click option'

Astrlen(str)

Bi

Cstrlen(dict[j])

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using length of the whole string instead of dictionary word length.

Using index variables i or j as length.

strncmp compares the first n characters; n should be length of the dictionary word to match substring correctly.

4fill in blank

hard

Fill both blanks to correctly update the DP array and loop index.

DSA C

if (dp[i] && strncmp(str + i, dict[j], [1]) == 0) {
    dp[i + [2]] = 1;
}

Drag options to blanks, or click blank then click option'

Astrlen(dict[j])

Bstrlen(str)

Ci

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different lengths for comparison and dp update.

Using string length or indices instead of dictionary word length.

We compare substring length with dictionary word length and update dp at position i + length of dict word.

5fill in blank

hard

Fill all three blanks to complete the main loop for the word break problem.

DSA C

for (int [1] = 0; [2] < len; [3]++) {
    for (int j = 0; j < dictSize; j++) {
        if (dp[i] && strncmp(str + i, dict[j], strlen(dict[j])) == 0) {
            dp[i + strlen(dict[j])] = 1;
        }
    }
}

Drag options to blanks, or click blank then click option'

Ai

Blen

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in loop parts.

Incorrect loop condition or increment.

The outer loop uses variable i from 0 to len-1, incrementing i each time.

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