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 (s.[1] == 0) return true;

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

ALength

BIsEmpty

CEquals("")

DIsNullOrEmpty

Attempts:

3 left

💡 Hint

Common Mistakes

Using IsNullOrEmpty which checks for null or empty, but here we only want to check empty.

Using a method that does not exist like IsEmpty.

The correct way to check if a string is empty in C# is by checking if its Length is 0.

2fill in blank

medium

Complete the code to iterate over all prefixes of the string.

DSA C#

for (int i = 1; i <= s.[1]; i++)

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

ASize

BCount

CLength

DLength()

Attempts:

3 left

💡 Hint

Common Mistakes

Using Length() with parentheses which causes a compile error.

Using Count or Size which are not valid for strings.

In C#, the length of a string is accessed using the Length property without parentheses.

3fill in blank

hard

Fix the error in substring extraction to get the prefix.

DSA C#

string prefix = s.[1](0, i);

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

Aslice

BSubstring

Csubstr

DGetRange

Attempts:

3 left

💡 Hint

Common Mistakes

Using substr or slice which are JavaScript methods, not C#.

Using GetRange which is for lists, not strings.

The correct method to get a substring in C# is Substring(startIndex, length).

4fill in blank

hard

Fill both blanks to check if the prefix is in the word dictionary and to recurse on the suffix.

DSA C#

if (wordDict.[1](prefix) && WordBreak(s.[2](i), wordDict)) return true;

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

AContains

BContainsKey

CSubstring

DSubstr

Attempts:

3 left

💡 Hint

Common Mistakes

Using ContainsKey which is for dictionaries, not lists.

Using Substr which is not a C# method.

Use Contains to check if the prefix is in the list, and Substring to get the suffix starting at index i.

5fill in blank

hard

Fill all three blanks to implement memoization using a dictionary.

DSA C#

if (memo.[1](s)) return memo[s];
memo.[3](s, [2]);
return memo[s];

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

AContainsKey

Btrue

Cfalse

DAdd

Attempts:

3 left

💡 Hint

Common Mistakes

Using Contains instead of ContainsKey for dictionaries.

Assigning values without adding to the dictionary.

Check if the memo dictionary contains the string key using ContainsKey. Store the boolean result true or false. Use Add to insert new entries.

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. The longest word in wordDict that appears in s

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

C. The number of unique characters in s that are also in wordDict

D. Whether s is a palindrome

Solution

1. Step 1: Understand the problem statement

   The Word Break Problem asks if the string s can be split into words all found in the dictionary wordDict.

2. Step 2: Identify the correct description

   Whether s can be segmented into a space-separated sequence of one or more dictionary words correctly describes this segmentation check, while others describe unrelated tasks.

3. Final Answer:

   Whether 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 checks if string splits into dictionary words [OK]

Common Mistakes:

• Confusing Word Break with palindrome checking
• Thinking it counts characters instead of segmenting
• Assuming it finds longest word instead of segmentation

2. Which of the following C# method signatures is correct for a function solving the Word Break Problem that returns a boolean indicating if the string can be segmented?

easy

A. public string WordBreak(int s, IList<string> wordDict)

B. public int WordBreak(string s, List<int> wordDict)

C. public bool WordBreak(string s, IList<string> wordDict)

D. public bool WordBreak(string s, string wordDict)

Solution

1. Step 1: Check parameter types for Word Break

   The input string s should be a string, and the dictionary wordDict should be a list of strings.

2. Step 2: Verify return type and parameters

   public bool WordBreak(string s, IList<string> wordDict) matches the expected signature: returns bool, takes string and IList<string>. Others have wrong types or return types.

3. Final Answer:

   public bool WordBreak(string s, IList<string> wordDict) -> Option C

4. Quick Check:

   Correct method signature = public bool WordBreak(string s, IList<string> wordDict) [OK]

Hint: Return bool, input string and list of strings for dictionary [OK]

Common Mistakes:

• Using wrong return type like int or string
• Passing dictionary as string instead of list
• Using wrong parameter types like int for string

3. What will be the output of the following C# code snippet?

var s = "leetcode";
var wordDict = new List<string> { "leet", "code" };
var result = WordBreak(s, wordDict);
Console.WriteLine(result);

bool WordBreak(string s, IList<string> wordDict) {
    var dp = new bool[s.Length + 1];
    dp[0] = true;
    for (int i = 1; i <= s.Length; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordDict.Contains(s.Substring(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.Length];
}

medium

A. False

B. Runtime exception

C. Compilation error

D. True

Solution

1. Step 1: Understand the dynamic programming logic

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

2. Step 2: Trace the substring checks for "leetcode"

   "leet" and "code" are in wordDict. The code finds "leet" at start and then "code" after, so dp[s.Length] becomes true.

3. Final Answer:

   True -> Option D

4. Quick Check:

   Substring segmentation found = True [OK]

Hint: Check dp array updates for substrings in dictionary [OK]

Common Mistakes:

• Assuming false because of substring indexing confusion
• Thinking code throws exception on substring
• Ignoring dp array meaning

4. Identify the bug in the following Word Break implementation and select the correct fix:

bool WordBreak(string s, IList<string> wordDict) {
    var dp = new bool[s.Length];
    dp[0] = true;
    for (int i = 1; i <= s.Length; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordDict.Contains(s.Substring(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[s.Length];
}

medium

A. Change dp array size to s.Length + 1 and initialize dp[0] = true

B. Change dp array to int[] instead of bool[]

C. Change substring parameters to s.Substring(i, j)

D. Remove the inner loop and check only j = 0

Solution

1. Step 1: Identify dp array size and indexing issue

   The dp array should have length s.Length + 1 to represent empty substring at dp[0]. Current dp size is s.Length, causing index errors.

2. Step 2: Fix dp initialization and size

   Changing dp to size s.Length + 1 and setting dp[0] = true fixes indexing and logic.

3. Final Answer:

   Change dp array size to s.Length + 1 and initialize dp[0] = true -> Option A

4. Quick Check:

   dp size and initialization fix = Change dp array size to s.Length + 1 and initialize dp[0] = true [OK]

Hint: dp array length = s.Length + 1 for empty substring base case [OK]

Common Mistakes:

• Using dp size s.Length causes index out of range
• Incorrect substring parameters
• Removing inner loop breaks logic

5. Given the string s = "applepenapple" and dictionary wordDict = ["apple", "pen"], which of the following modifications to the Word Break algorithm will correctly count the number of ways to segment s into dictionary words instead of just checking if segmentation is possible?

hard

A. Change dp array to int[], initialize dp[0] = 1, and update dp[i] += dp[j] when substring is in dictionary

B. Use a boolean dp array but return dp[s.Length] as integer count

C. Use recursion without memoization to count ways

D. Change dp array to string[] to store segmented sentences

Solution

1. Step 1: Understand difference between boolean and count dp

   Boolean dp tracks if segmentation is possible. To count ways, dp must store counts as integers.

2. Step 2: Modify dp to int[], initialize dp[0] = 1, and accumulate counts

   When substring s[j..i-1] is in dictionary, add dp[j] to dp[i] to count all ways.

3. Final Answer:

   Change dp array to int[], initialize dp[0] = 1, and update dp[i] += dp[j] when substring is in dictionary -> Option A

4. Quick Check:

   Counting ways requires int dp and accumulation [OK]

Hint: Use int dp and add counts for each valid substring [OK]

Common Mistakes:

• Using boolean dp for counting ways
• Returning boolean as integer count
• Ignoring memoization in recursion