<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 solve(string s, vector<string>& wordDict, int i, vector<bool>& dp){
        if(i==s.size()) return true;
        if(i>s.size()) return false;

        if(dp[i]!=false) return true;

        bool ans=false;
        for(auto word : wordDict){
            int n=word.size();
            if(i+n<=s.size()){
                string currword = s.substr(i, n);
                if(currword == word){
                    bool a = solve(s, wordDict, i+n, dp);
                    if(a || ans) ans=true;
                }
            }
        }
        return dp[i]=ans;
    }

    bool wordBreak(string s, vector<string>& wordDict) {
        vector<bool> dp(s.size()+1, false);
        return solve(s, wordDict, 0, dp);
    }
*/

    bool wordBreak(string s, vector<string>& wordDict) {
        int z = s.size();
        vector<bool> dp(z+z, false);
        dp[z]=true;

        for(int i=z-1; i>=0; i--){
            bool ans=false;
            for(auto word : wordDict){
                int n=word.size();
                if(i+n<=s.size()){
                    string currword = s.substr(i, n);
                    if(currword == word){
                        bool a = dp[i+n];
                        if(a || ans) ans=true;
                    }
                }
            }
        dp[i]=ans;
        }
    return dp[0];
    }
};
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