<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, int j, vector<vector<int>>& dp) {
        if (i > j || i >= s.length() || j >= s.length()) {
            return true;
        }
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        bool ans = false;
        string st = s.substr(i, j - i + 1);
        for (string str : wordDict) {
            auto it = st.find(str);
            if (it != string::npos) {
                bool split = solve(s, wordDict, i, i + it - 1, dp) && solve(s, wordDict, i + it + str.length(), j, dp);
                ans |= split;
            }
        }
        return dp[i][j] = ans;
    }
    bool wordBreak(string s, vector<string>& wordDict) {
        vector<vector<int>> dp(s.length() + 1, vector<int>(s.length() + 1, -1));
        return solve(s, wordDict, 0, s.length() - 1, dp);
    }
};
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