LeetCode in Cpp

139. Word Break

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.

Solution

#include <string>
#include <vector>
#include <unordered_set>
using namespace std;

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> set;
        int maxLen = 0;
        vector<bool> flag(s.length() + 1, false);

        for (const string& word : wordDict) {
            set.insert(word);
            maxLen = max(maxLen, (int)word.length());
        }

        for (int i = 1; i <= maxLen; ++i) {
            if (dfs(s, 0, i, maxLen, set, flag)) {
                return true;
            }
        }
        return false;
    }

private:
    bool dfs(const string& s, int start, int end, int maxLen, unordered_set<string>& set, vector<bool>& flag) {
        if (!flag[end] && set.count(s.substr(start, end - start))) {
            flag[end] = true;
            if (end == s.length()) {
                return true;
            }
            for (int i = 1; i <= maxLen; ++i) {
                if (end + i <= s.length() && dfs(s, end, end + i, maxLen, set, flag)) {
                    return true;
                }
            }
        }
        return false;
    }
};

This site is open source. Improve this page.