Word Break
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
Note:
- The same word in the dictionary may be reused multiple times in the segmentation.
- You may assume the dictionary does not contain duplicate words.
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
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int n=s.size();
unordered_map<string,int> mp;
int l=0;
for(auto w:wordDict)
{
mp[w]++;
int k=w.size();
l=max(l,k);
}
vector<bool> dp(n+1,false);
dp[0]=true;
for(int i=1;i<=n;++i)
{
for(int j=0;j<i;++j)
{
if(!dp[j]||(i-j>l))continue;
if(mp.find(s.substr(j,i-j))!=mp.end())
{
dp[i]=true;
break;
}
}
}
return dp[n];
}
};
