Why Word Break Problem in DSA Typescript?

The Big Idea

What if you could instantly know if a jumble of letters forms real words without guessing every split?

The Scenario

Imagine you have a long string of letters without spaces, like a secret message, and you want to find out if it can be split into meaningful words from a dictionary.

Doing this by checking every possible split manually is like trying to guess where to put spaces in a long sentence without any clues.

The Problem

Manually trying all ways to split the string is very slow and confusing because the number of ways grows very fast as the string gets longer.

You might miss some splits or repeat the same checks many times, making it frustrating and error-prone.

The Solution

The Word Break Problem uses a smart way to remember which parts of the string can be split successfully, so it doesn't waste time checking the same parts again.

This method quickly tells you if the whole string can be broken into dictionary words without trying every possibility from scratch.

Before vs After

✗ Before

function canBreak(s: string, wordDict: string[]): boolean {
  if (s.length === 0) return true;
  for (let i = 1; i <= s.length; i++) {
    let prefix = s.substring(0, i);
    if (wordDict.includes(prefix) && canBreak(s.substring(i), wordDict)) {
      return true;
    }
  }
  return false;
}

✓ After

function canBreak(s: string, wordDict: string[]): boolean {
  const dp: boolean[] = Array(s.length + 1).fill(false);
  dp[0] = true;
  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && wordDict.includes(s.substring(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }
  return dp[s.length];
}

What It Enables

This concept enables fast and reliable checking if a string can be split into valid words, making text processing and language tasks much easier.

Real Life Example

When you type a sentence without spaces on your phone, the system can quickly figure out the correct words to display by solving the Word Break Problem behind the scenes.

Key Takeaways

Manual splitting is slow and repeats work.

Dynamic programming remembers results to speed up checks.

Useful for text processing and language understanding.

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 can be reversed to form a word in wordDict

B. If s contains any substring that is not in wordDict

C. If s is exactly equal to one of the words in wordDict

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

Solution

1. Step 1: Understand the problem goal

   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: Clarify what is being checked

   It is not about substrings not in the dictionary or exact matches only, but about segmenting s fully into dictionary words.

3. Final Answer:

   If s can be segmented into a sequence of dictionary words -> Option D

4. Quick Check:

   Word Break = segment string into dictionary words [OK]

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

Common Mistakes:

• Thinking it checks for partial substrings only
• Confusing exact match with segmentation
• Assuming reversed words are considered

2. Which of the following TypeScript function signatures correctly represents a function solving the Word Break Problem?

easy

A. function wordBreak(words: string[], s: string): number

B. function wordBreak(s: string, wordDict: string[]): boolean

C. function wordBreak(s: number, wordDict: string): string

D. function wordBreak(s: string, wordDict: string): void

Solution

1. Step 1: Identify correct parameter types

   The function should take a string s and an array of strings wordDict representing the dictionary words.

2. Step 2: Identify correct return type

   The function returns a boolean indicating if s can be segmented or not.

3. Final Answer:

   function wordBreak(s: string, wordDict: string[]): boolean -> Option B

4. Quick Check:

   Correct params and boolean return [OK]

Hint: Function takes string and string array, returns boolean [OK]

Common Mistakes:

• Using wrong parameter types
• Returning wrong data type
• Swapping parameter order

3. What will be the output of the following TypeScript code?

function wordBreak(s: string, wordDict: string[]): boolean {
  const wordSet = new Set(wordDict);
  const dp = Array(s.length + 1).fill(false);
  dp[0] = true;
  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && wordSet.has(s.substring(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }
  return dp[s.length];
}

console.log(wordBreak("leetcode", ["leet", "code"]));

medium

A. SyntaxError

B. false

C. true

D. undefined

Solution

1. Step 1: Understand the dynamic programming approach

   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 example input "leetcode"

   "leet" and "code" are in the dictionary. The code finds "leet" at start and sets dp[4] true, then "code" from index 4 to 8 sets dp[8] true. So final dp[s.length] is true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   DP finds full segmentation = true [OK]

Hint: Check dp array for full string length true [OK]

Common Mistakes:

• Confusing substring indices
• Not breaking inner loop after true
• Misunderstanding dp array meaning

4. Identify the bug in this TypeScript code for the Word Break Problem:

function wordBreak(s: string, wordDict: string[]): boolean {
  const wordSet = new Set(wordDict);
  const dp = Array(s.length).fill(false);
  dp[0] = true;
  for (let i = 1; i <= s.length; i++) {
    for (let j = 0; j < i; j++) {
      if (dp[j] && wordSet.has(s.substring(j, i))) {
        dp[i] = true;
        break;
      }
    }
  }
  return dp[s.length];
}

medium

A. dp array size is incorrect; should be s.length + 1

B. wordSet should be an array, not a Set

C. The substring indices are reversed in s.substring(j, i)

D. dp[0] should be false initially

Solution

1. Step 1: Check dp array initialization

   The dp array is created with size s.length, but it needs to be s.length + 1 to include the base case dp[0].

2. Step 2: Understand consequences of wrong dp size

   Accessing dp[s.length] will be out of bounds, causing runtime errors or incorrect results.

3. Final Answer:

   dp array size is incorrect; should be s.length + 1 -> Option A

4. Quick Check:

   dp size must cover 0 to s.length indices [OK]

Hint: dp array length = string length + 1 [OK]

Common Mistakes:

• Using dp array of wrong size
• Confusing Set with array for dictionary
• Incorrect substring index order

5. Given the string s = "applepenapplepenapple" and dictionary wordDict = ["apple", "pen"], what is the minimum number of segments needed to fully segment s into dictionary words?

hard

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Break down the string into dictionary words

   The string is "applepenapplepenapple". The dictionary words are "apple" and "pen". The string can be segmented as "apple" + "pen" + "apple" + "pen" + "apple".

2. Step 2: Count the segments

   There are 5 segments total: apple, pen, apple, pen, apple.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Count dictionary words in segmentation = 5 [OK]

Hint: Count dictionary words in full segmentation [OK]

Common Mistakes:

• Counting characters instead of words
• Missing repeated segments
• Assuming fewer segments possible