# 139. Word Break | LeetCode Solution

## 139. Word Break | LeetCode Solution

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.

class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
vector<long long int>dp(s.length(),0);
unordered_map<string,int>m;
for(auto i:wordDict)m[i]=1;

for(int i=0; i<s.length(); i++){
int curr=0;
for(int j=0; j<=i; j++){
// cout<<s.substr(j,i-j+1)<<endl;
if(m.find(s.substr(j,i-j+1))!=m.end()){
if(j>0)
curr+=dp[j-1];
else if(j==0)curr+=1;
}
}
// cout<<curr<<endl;
dp[i]=curr;
}
for(auto i:dp)cout<<i<<" ";
return dp[s.length()-1]>0;
}
};