<p>Given a string <code>s</code> and a dictionary of strings <code>wordDict</code>, return <code>true</code> if <code>s</code> can be segmented into a space-separated sequence of one or more dictionary words.</p>

<p><strong>Note</strong> that the same word in the dictionary may be reused multiple times in the segmentation.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;, wordDict = [&quot;leet&quot;,&quot;code&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> Return true because &quot;leetcode&quot; can be segmented as &quot;leet code&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;applepenapple&quot;, wordDict = [&quot;apple&quot;,&quot;pen&quot;]
<strong>Output:</strong> true
<strong>Explanation:</strong> Return true because &quot;applepenapple&quot; can be segmented as &quot;apple pen apple&quot;.
Note that you are allowed to reuse a dictionary word.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;catsandog&quot;, wordDict = [&quot;cats&quot;,&quot;dog&quot;,&quot;sand&quot;,&quot;and&quot;,&quot;cat&quot;]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 300</code></li>
	<li><code>1 &lt;= wordDict.length &lt;= 1000</code></li>
	<li><code>1 &lt;= wordDict[i].length &lt;= 20</code></li>
	<li><code>s</code> and <code>wordDict[i]</code> consist of only lowercase English letters.</li>
	<li>All the strings of <code>wordDict</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        //unordered_map of the wordDict
        unordered_map<string,bool> dict;
        for(int i=0;i<wordDict.size();i++){
            dict[wordDict[i]] = true;
        }
        int size = s.length();
        vector<vector<bool>> dp;
        for(int i=0;i<size;i++){
            vector<bool> kemp;
            for(int j=0;j<size;j++){
                kemp.push_back(false);
            }
            dp.push_back(kemp);
        }
        for(int length=1;length<=size;length++){
            for(int i=0;i<size-length+1;i++){
                int j = i+length-1;
                cout<<length<<" "<<i<<" "<<j<<endl;
                string temp;
                for(int k=i;k<=j;k++){
                    temp.push_back(s[k]);
                }
                if(dict.find(temp)!=dict.end()){
                    dp[i][j] = true;
                }else{
                    for(int k=i;k<j;k++){
                        bool left = dp[i][k];
                        bool right = dp[k+1][j];
                        if(left&&right){
                            dp[i][j] = true;
                            break;
                        }
                    }
                }
            }
        }
        cout<<endl;
        for(int i=0;i<size;i++){
            for(int j=0;j<size;j++){
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }
        return dp[0][size-1];
    }

    bool wordBreakHelper(int startIndex,int endIndex, unordered_map<string,bool> &dict,string s){   
        if(startIndex>endIndex){
            return true;
        }
        string temp;
        for(int i=startIndex;i<=endIndex;i++){
            temp.push_back(s[i]);
        }
        if(dict.find(temp)!=dict.end()){
            return true;
        }else{
            for(int k=startIndex;k<endIndex;k++){
            bool left = wordBreakHelper(startIndex,k,dict,s);
            bool right = wordBreakHelper(k+1,endIndex,dict,s);
            if(left&&right){
                return true;
            }
        }
        }
        return false;
    }
};
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