<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:
    vector<int> dp;
    
    bool check(int level, int n, string s, unordered_map<string, int> &mp) {
        // Base case: if the level is below 0, the word can be broken successfully
        if (level < 0) return true;
        
        // Cache check: if already computed, return cached value
        if (dp[level] != -1) return dp[level];
        
        string ns = "";
        bool ans = false;
        
        // Try to split the string by checking all possible substrings
        for (int i = level; i >= 0; i--) {
            ns = s[i] + ns;
            // If substring ns is in dictionary and the remaining can be split
            if (mp[ns] == 1 && check(i - 1, n, s, mp)) {
                ans = true;
                break;
            }
        }
        
        // Save and return the result
        return dp[level] = ans;
    }

    bool wordBreak(string s, vector<string>& wordDict) {
        unordered_map<string, int> mp;
        int n = s.length();
        dp.assign(n, -1);
        
        // Insert all dictionary words into the map
        for (auto &v : wordDict) {
            mp[v] = 1;
        }
        
        // Start the recursion from the last index of the string
        return check(n - 1, n, s, mp);
    }
};

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