Word Break Problem in DSA C

Mental Model

We want to check if a string can be split into valid words from a dictionary. We try breaking the string at every position and see if the left part is a word and the right part can also be split.

Analogy: Imagine you have a long sentence without spaces and a list of known words. You try to put spaces in different places to see if the sentence can be made up of those known words.

string: w o r d b r e a k
dictionary: [word, break, problem]

Initial:
"wordbreak"

Try splits:
[word] + [break]
[wordb] + [reak]
[wordbr] + [eak]
...

Dry Run Walkthrough

Input: string: "wordbreak", dictionary: ["word", "break", "problem"]

Goal: Determine if "wordbreak" can be segmented into dictionary words

Step 1: Check prefix "w" - not in dictionary, move to next prefix

"w" (not word), continue checking prefixes

Step 2: Check prefix "wo" - not in dictionary, move to next prefix

"wo" (not word), continue checking prefixes

Step 3: Check prefix "wor" - not in dictionary, move to next prefix

"wor" (not word), continue checking prefixes

Step 4: Check prefix "word" - found in dictionary, now check suffix "break"

"word" (word), suffix "break" to check

Step 5: Check suffix "break" - found in dictionary, so "wordbreak" can be segmented

"word" + "break" both words, success

Result:

"word" -> "break" -> null (string can be segmented)

Annotated Code

DSA C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

// Check if word is in dictionary
bool isInDict(const char* word, const char* dict[], int dictSize) {
    for (int i = 0; i < dictSize; i++) {
        if (strcmp(word, dict[i]) == 0) {
            return true;
        }
    }
    return false;
}

// Word Break function using recursion and memoization
bool wordBreakHelper(const char* s, const char* dict[], int dictSize, int start, int memo[]) {
    int len = strlen(s);
    if (start == len) {
        return true; // reached end successfully
    }
    if (memo[start] != -1) {
        return memo[start]; // return cached result
    }
    for (int end = start + 1; end <= len; end++) {
        int subLen = end - start;
        char sub[subLen + 1];
        strncpy(sub, s + start, subLen);
        sub[subLen] = '\0';
        if (isInDict(sub, dict, dictSize)) {
            if (wordBreakHelper(s, dict, dictSize, end, memo)) {
                memo[start] = 1;
                return true;
            }
        }
    }
    memo[start] = 0;
    return false;
}

bool wordBreak(const char* s, const char* dict[], int dictSize) {
    int len = strlen(s);
    int memo[len];
    for (int i = 0; i < len; i++) {
        memo[i] = -1; // -1 means not computed
    }
    return wordBreakHelper(s, dict, dictSize, 0, memo);
}

int main() {
    const char* dict[] = {"word", "break", "problem"};
    const char* s = "wordbreak";
    if (wordBreak(s, dict, 3)) {
        printf("word break possible\n");
    } else {
        printf("word break not possible\n");
    }
    return 0;
}

if (start == len) { return true; }

base case: reached end means successful segmentation

if (memo[start] != -1) { return memo[start]; }

use memo to avoid repeated work

for (int end = start + 1; end <= len; end++) { ... }

try all prefixes from current start

if (isInDict(sub, dict, dictSize)) { if (wordBreakHelper(...)) { memo[start] = true; return true; } }

if prefix is word and suffix can be segmented, mark true and return

memo[start] = false; return false;

no valid segmentation found from this start

Output

word break possible

Complexity Analysis

Time: O(n^3) because for each start index we try all end indices and substring copying takes O(n)

Space: O(n) for memoization array to store results for each start index

vs Alternative: Naive recursion without memoization is exponential because it recomputes subproblems many times

Edge Cases

empty string

returns true because empty string can be segmented trivially

DSA C

if (start == len) { return true; }

string with no matching dictionary words

returns false as no segmentation possible

DSA C

memo[start] = false; return false;

string equals a dictionary word

returns true immediately

DSA C

if (start == len) { return true; }

When to Use This Pattern

When you need to check if a string can be split into valid dictionary words, use the Word Break pattern with recursion and memoization to efficiently explore all splits.

Common Mistakes

Mistake: Not using memoization causing exponential time and timeout
Fix: Add a memo array to store results for each start index and reuse them

Mistake: Not checking the base case when start reaches string length
Fix: Add base case: if start == length, return true

Mistake: Incorrect substring extraction causing wrong comparisons
Fix: Use strncpy carefully and null-terminate substring before dictionary check

Summary

Checks if a string can be split into dictionary words by trying all prefixes and recursively checking suffixes.

Use when you want to validate if a string can be segmented into known words.

The key insight is to try every prefix and remember results to avoid repeated work.

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