Interview Classic 150 Questions P139 Word Break

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Timeline

Timeline

2025-12-13

init

Dynamic Programming

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Problem:

P139 Word Break

https://leetcode.cn/problems/word-break/?envType=study-plan-v2&envId=top-interview-150

dp[i] indicates whether the prefix s[0…i-1] can be successfully split into words from the dictionary. dp[0] = true, meaning the empty string can be split.
dp[The condition for i] to be true is that dp[j](0 <= j < i)is true, and s.substr(j, i-j) is a word in wordDict.

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

class Solution {
    public:
        bool wordBreak(string s, vector<string> &wordDict)
        {
                int i, j, n = s.size();
                unordered_set<string> dict(wordDict.begin(), wordDict.end());
                // dp[i] indicates whether the prefix s[0..i-1] can be successfully split
                vector<bool> dp(n + 1, false);
                dp[0] = true; // The empty string can be split
                for (i = 1; i <= n; i++) {
                        for (j = 0; j < i; j++) { // s[0..i-1]
                                //
                                if (dp[j] && dict.count(s.substr(j, i - j))) {
                                        dp[i] = true;
                                        break;
                                }
                        }
                }
                return dp[n];
        }
};

LeetCode Hot 100 rewrite: used prefix tree + BFS approach

#include <string>
#include <vector>
#include <queue>

using std::vector;
using std::string;
using std::queue;

struct TrieNode {
        vector<TrieNode *> children;
        char data;
        bool is_end;

        TrieNode()
        {
                this->children = vector<TrieNode *>(26, nullptr);
                this->is_end = false;
        }
};

class Solution {
    public:
        bool wordBreak(string s, vector<string> &wordDict)
        {
                // 1 <= s.length <= 300
                // 1 <= wordDict.length <= 1000
                // 1 <= wordDict[i].length <= 20
                // s and wordDict[i] consist only of lowercase English letters.
                // All strings in wordDict are distinct.

                TrieNode *root = new TrieNode, *p;
                int i, n = s.size();

                for (string word : wordDict) {
                        p = root;

                        for (char ch : word) {
                                if (p->children[ch - 'a'] == nullptr)
                                        p->children[ch - 'a'] = new TrieNode;

                                p = p->children[ch - 'a'];
                                p->data = ch;
                        }
                        p->is_end = true;
                }

                queue<int> que;
                vector<bool> visited(n, false);
                que.push(0);
                visited[0] = true;

                while (!que.empty()) {
                        int start = que.front();
                        que.pop();
                        if (start == n)
                                return true;

                        p = root;
                        for (i = start; i < n; i++) {
                                if (p->children[s[i] - 'a'] == nullptr)
                                        break;

                                p = p->children[s[i] - 'a'];

                                if (p->is_end && !visited[i + 1]) {
                                        que.push(i + 1);
                                        visited[i + 1] = true;
                                }
                        }
                }

                return false;
        }
};