139. Word Break

Problem Statement

Medium

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.

C++

Source file

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        vector<bool> dp(s.length());
        for (int i = 0; i < s.length(); i++) {
            for (string word: wordDict) {
                // Handle out of bounds case
                if (i < word.length() - 1) {
                    continue;
                }
                
                if (i == word.length() - 1 || dp[i - word.length()]) {
                    if (s.substr(i - word.length() + 1, word.length()) == word) {
                        dp[i] = true;
                        break;
                    }
                }
            }
        }
        
        return dp[s.length() - 1];
    }
};