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;
    }
};