<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:

    int dp[305][305];

    bool solve(int cur_index, int last_index, map <string, int>& freq, string& s) {
        int n = s.size();

        if (cur_index == n && last_index == n-1) {

            // cout << "TRUEEEEE!!!" << endl;

            return true;
        }

        if (cur_index == n) {
            
            // cout << "FAAALLLSE!!!   last_idx = " << last_index << endl;

            return false;
        }

        if (last_index != -1 && dp[cur_index][last_index] != -1) {

            return dp[cur_index][last_index];
        }

        string word = "";
        for (int i=last_index+1 ; i<=cur_index ; i++) {

            word += s[i];
        }

        // cout << "last_idx+1 = " << last_index+1 << ", cur_index = " << cur_index << ", word = " << word << endl;

        if (freq.find(word) != freq.end()) {

            // cout << "Word is present!!" << endl;

            bool ret = solve(cur_index+1, cur_index, freq, s);

            if (ret) {

                if (last_index != -1) {

                    return dp[cur_index][last_index] = 1;
                }
                
                return ret;
            }
        }

        if (last_index != -1) {

            return dp[cur_index][last_index] = solve(cur_index+1, last_index, freq, s);
        }

        return solve(cur_index+1, last_index, freq, s);
    }

    bool wordBreak(string s, vector<string>& wordDict) {

        map <string, int> freq;

        for (int i=0 ; i<wordDict.size() ; i++) {

            freq[wordDict[i]]++;
        }

        for (int i=0 ; i<305; i++) {

            for (int j=0 ; j<305 ; j++) {

                dp[i][j] = -1;
            }
        }

        return solve(0, -1, freq, s);
    }
};
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