Word Break

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = “leetcode”, wordDict = [“leet”,“code”] Output: true Explanation: Return true because “leetcode” can be segmented as “leet code”.

Example 2:

Input: s = “applepenapple”, wordDict = [“apple”,“pen”] Output: true Explanation: Return true because “applepenapple” can be segmented as “apple pen apple”. Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = “catsandog”, wordDict = [“cats”,“dog”,“sand”,“and”,“cat”] Output: false

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Approach: Dynamic Programming (1D)

Algorithm

1. Let dp[i] indicate whether the prefix s[0..i-1] can be segmented into dictionary words
2. dp[0] = true (empty prefix)
3. For each end index i, check every word: if dp[i - len] is true and s.substring(i-len, i) equals the word, set dp[i] = true
4. Return dp[s.length()]

Time & Space Complexity

• Time Complexity: O(n * m * L) - n is the string length, m is the dictionary size, and each substring comparison costs O(L)
• Space Complexity: O(n) - the DP array

Java Implementation

import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class WordBreak {

    /**
     * Check whether s can be segmented into dictionary words.
     * @param s Input string
     * @param wordDict Dictionary of words
     * @return true if s is segmentable
     */
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> wordSet = new HashSet<>(wordDict);
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;

        for (int i = 1; i <= s.length(); i++) {
            for (String word : wordDict) {
                int len = word.length();
                if (i >= len && dp[i - len]
                        && wordSet.contains(s.substring(i - len, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[s.length()];
    }

    // Test method
    public static void main(String[] args) {
        WordBreak wb = new WordBreak();
        System.out.println("Input: s = \"leetcode\", wordDict = [\"leet\",\"code\"]");
        System.out.println("Output: " + wb.wordBreak("leetcode", List.of("leet", "code"))); // Expected: true

        System.out.println("Input: s = \"catsandog\", wordDict = [\"cats\",\"dog\",\"sand\",\"and\",\"cat\"]");
        System.out.println("Output: " + wb.wordBreak("catsandog",
                List.of("cats", "dog", "sand", "and", "cat"))); // Expected: false
    }
}

Alternative Approach: DP over Prefixes

public boolean wordBreakPrefix(String s, List<String> wordDict) {
    Set<String> wordSet = new HashSet<>(wordDict);
    boolean[] dp = new boolean[s.length() + 1];
    dp[0] = true;

    for (int i = 1; i <= s.length(); i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && wordSet.contains(s.substring(j, i))) {
                dp[i] = true;
                break;
            }
        }
    }

    return dp[s.length()];
}

Example Walkthrough

For s = "leetcode", wordDict = ["leet","code"]:

1. dp[0] = true
2. At i=4, substring(0,4)=“leet” is in the set and dp[0] is true -> dp[4] = true
3. At i=8, substring(4,8)=“code” is in the set and dp[4] is true -> dp[8] = true
4. Return true

For s = "catsandog", no combination of the dictionary words covers the full string, so dp[8] stays false.

Key Points

1. Overlapping Subproblems: Whether a prefix is segmentable is reused by later positions
2. dp[0] = true: The empty prefix seeds the recursion
3. Word Reuse: The dictionary set allows a word to be used any number of times
4. Early Break: Once dp[i] is set, no further words need checking
5. Two Variants: Iterating over words or over prefix split points both run in polynomial time