<p>Given a string <code>s</code>, return <em>the number of <strong>palindromic substrings</strong> in it</em>.</p>

<p>A string is a <strong>palindrome</strong> when it reads the same backward as forward.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within the string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> Three palindromic strings: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaa&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> Six palindromic strings: &quot;a&quot;, &quot;a&quot;, &quot;a&quot;, &quot;aa&quot;, &quot;aa&quot;, &quot;aaa&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:

    
    vector<string> all_palindromes;
    unordered_set<string> distinct_palindromes;

    int countSubstrings(string s) {
        int n = s.length();
        int ans = 0;

        int even = 0;
        int odd = 0;

        for(int i = 0; i<n; ++i){
            odd = palindromeCount(s,i,i);
            even = palindromeCount(s,i,i+1);
            ans += odd + even;
        }
        
        //cout << " All palindromes :"<< "\n";
        //for(const auto &s : all_palindromes){
        //    cout << s << "\n";
        //}

        cout << " Distinct palindromes :"<< "\n";
        for(const auto &ds : distinct_palindromes){
            cout << ds << "\n";
        }

        return ans;
        
    }

    int palindromeCount(const string &s, int left, int right){
        int count = 0;
        while(left >=0 && right < s.length() && s[left] == s[right]){
            //all_palindromes.push_back(s.substr(left, right - left + 1));
            distinct_palindromes.insert(s.substr(left, right - left + 1));
            left--;
            right++;
            count++;

            
        }
        return count;
    }
};

//TC : O(N^2), where N is length of string
//SC : O(1)
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