<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 {

private:
    int solve(int ind, set<string>& st, string& s, vector<int>& dp) {
        if(ind == s.size()) return 1;

        if(dp[ind] != -1) return dp[ind];

        int ans = 0;

        for(int i = ind; i < s.size(); i++) {
            string temp = s.substr(ind, i-ind+1);

            if(st.find(temp) != st.end()) {
                ans |= solve(i+1, st, s, dp);
            }
        }

        return ans;

    }

public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = wordDict.size();

        vector<int> dp(s.size()+1, 0);

        set<string> st(wordDict.begin(), wordDict.end());

        // bool ans = solve(0, st, s, dp);
        dp[s.size()] = 1;

        for(int i = s.size()-1; i >= 0; i--) {
            for(auto word : wordDict) {
                if(i + word.size() <= s.size() && s.substr(i, word.size()) == word) {
                    dp[i] = dp[i+word.size()];
                }

                if(dp[i]) break;
            }
        }

        if(dp[0] == 1) return true;
        return false;
    }
};
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