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.