Word Break Problem

Problem

Given a string and a dictionary of words, determine if the string can be segmented into a sequence of dictionary words.

Input:  s = "leetcode", wordDict = ["leet", "code"]
Output: true

Python Solution

def word_break(s, word_dict):
    words = set(word_dict)
    n = len(s)
    dp = [False] * (n + 1)
    dp[0] = True
    for i in range(1, n + 1):
        for j in range(i):
            if dp[j] and s[j:i] in words:
                dp[i] = True
                break
    return dp[n]

C++ Solution

bool wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> words(wordDict.begin(), wordDict.end());
    int n = s.size();
    vector<bool> dp(n + 1, false);
    dp[0] = true;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] && words.count(s.substr(j, i - j))) {
                dp[i] = true;
                break;
            }
        }
    }
    return dp[n];
}

Java Solution

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> words = new HashSet<>(wordDict);
        int n = s.length();
        boolean[] dp = new boolean[n + 1];
        dp[0] = true;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                if (dp[j] && words.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[n];
    }
}

Complexity

Time: O(n² ) for the DP loops, plus O(k) average per substring lookup in the word set, where n is the string length. Space: O(n) for the DP array, plus O(w) for the dictionary set of w words.