<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, int start, unordered_set<string>& wordSet, vector<int>& memo){

        // Base case: if we have processed the entire string
        if (start == s.length()) {
            return true;
        }

        // If the current start index has been processed before, return the memoized result
        if (memo[start] != -1) {
            return memo[start];
        }

        // Try every possible end point of the substring
        for (int end = start + 1; end <= s.length(); end++) {
            // Check if s[start...end] is in the dictionary
            if (wordSet.find(s.substr(start, end - start)) != wordSet.end()) {
                // If the substring is found, recursively check for the remaining part of the string
                if (solve(s, end, wordSet, memo)) {
                    return memo[start] = true; // Store the result in the memo table
                }
            }
        }

        // If no valid segmentation found, mark it as false in the memo table
        return memo[start] = false;
    }

    bool wordBreak(string s, vector<string>& wordDict) {
        
        //using set, to make O(1) lookups
        unordered_set<string> wordSet(wordDict.begin(), wordDict.end());

        //memoization array to store results (to avoid recomputing)
        vector<int> memo(s.length(), -1);

        return solve(s, 0, wordSet, memo);
    }
};
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