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# Word Break LeetCode Solution – Queslers

## Problem – Word Break

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.

### Word Break LeetCode Solution in Java

``````public class Solution {
public boolean wordBreak(String s, Set<String> dict) {

boolean[] f = new boolean[s.length() + 1];

f = true;

/* First DP
for(int i = 1; i <= s.length(); i++){
for(String str: dict){
if(str.length() <= i){
if(f[i - str.length()]){
if(s.substring(i-str.length(), i).equals(str)){
f[i] = true;
break;
}
}
}
}
}*/

//Second DP
for(int i=1; i <= s.length(); i++){
for(int j=0; j < i; j++){
if(f[j] && dict.contains(s.substring(j, i))){
f[i] = true;
break;
}
}
}

return f[s.length()];
}
}
``````

### Word Break LeetCode Solution in C++

``````bool wordBreak(string s, unordered_set<string> &dict) {
if(dict.size()==0) return false;

vector<bool> dp(s.size()+1,false);
dp=true;

for(int i=1;i<=s.size();i++)
{
for(int j=i-1;j>=0;j--)
{
if(dp[j])
{
string word = s.substr(j,i-j);
if(dict.find(word)!= dict.end())
{
dp[i]=true;
break; //next i
}
}
}
}

return dp[s.size()];
}
``````

### Word Break LeetCode Solution in Python

``````def word_break(s, words):
d = [False] * len(s)
for i in range(len(s)):
for w in words:
if w == s[i-len(w)+1:i+1] and (d[i-len(w)] or i-len(w) == -1):
d[i] = True
return d[-1]``````
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