Word Break Problem

Description

LeetCode Problem 139.

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:

``````1
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Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".
``````

Example 2:

``````1
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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:

``````1
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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.

Sample C++ Code

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class Solution {
public:
struct TrieNode {
TrieNode* arr[26];
bool is_end;
TrieNode() {
for (int i = 0; i < 26; i ++) {
arr[i] = nullptr;
}
is_end = false;
}
};

void insert(TrieNode* root, string s) {
TrieNode* curr = root;
for (int i = 0; i < s.size(); i ++) {
int idx = s[i] - 'a';
if (curr->arr[idx] == nullptr)
curr->arr[idx] = new TrieNode();
curr = curr->arr[idx];
}
curr->is_end = true;
}

bool wordBreak(string s, vector<string>& wordDict) {
TrieNode* root = new TrieNode();
for (int i = 0; i < wordDict.size(); i ++) {
insert(root, wordDict[i]);
}

int l = s.size();
vector<int> dp(l+1, 0);
dp[0] = 1;

for (int i = 0; i < l; i ++) {
if (dp[i] == 0)
continue;
TrieNode* curr = root;
for (int j = i; j < l; j ++) {
if (curr->arr[s[j]-'a'] == NULL)
break;
curr = curr->arr[s[j]-'a'];
if (curr->is_end == true)
dp[j+1] = 1;
}
}

return dp[l];
}
};
``````