<p>Given a string <code>s</code> of lowercase letters, you need to find the maximum number of <strong>non-empty</strong> substrings of <code>s</code> that meet the following conditions:</p>

<ol>
	<li>The substrings do not overlap, that is for any two substrings <code>s[i..j]</code> and <code>s[x..y]</code>, either <code>j &lt; x</code> or <code>i &gt; y</code> is true.</li>
	<li>A substring that contains a certain character <code>c</code> must also contain all occurrences of <code>c</code>.</li>
</ol>

<p>Find <em>the maximum number of substrings that meet the above conditions</em>. If there are multiple solutions with the same number of substrings, <em>return the one with minimum total length. </em>It can be shown that there exists a unique solution of minimum total length.</p>

<p>Notice that you can return the substrings in <strong>any</strong> order.</p>

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

<pre>
<strong>Input:</strong> s = &quot;adefaddaccc&quot;
<strong>Output:</strong> [&quot;e&quot;,&quot;f&quot;,&quot;ccc&quot;]
<b>Explanation:</b>&nbsp;The following are all the possible substrings that meet the conditions:
[
&nbsp; &quot;adefaddaccc&quot;
&nbsp; &quot;adefadda&quot;,
&nbsp; &quot;ef&quot;,
&nbsp; &quot;e&quot;,
  &quot;f&quot;,
&nbsp; &quot;ccc&quot;,
]
If we choose the first string, we cannot choose anything else and we&#39;d get only 1. If we choose &quot;adefadda&quot;, we are left with &quot;ccc&quot; which is the only one that doesn&#39;t overlap, thus obtaining 2 substrings. Notice also, that it&#39;s not optimal to choose &quot;ef&quot; since it can be split into two. Therefore, the optimal way is to choose [&quot;e&quot;,&quot;f&quot;,&quot;ccc&quot;] which gives us 3 substrings. No other solution of the same number of substrings exist.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abbaccd&quot;
<strong>Output:</strong> [&quot;d&quot;,&quot;bb&quot;,&quot;cc&quot;]
<b>Explanation: </b>Notice that while the set of substrings [&quot;d&quot;,&quot;abba&quot;,&quot;cc&quot;] also has length 3, it&#39;s considered incorrect since it has larger total length.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    inline vector<string> maxNumOfSubstrings(const std::string& s) noexcept {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        int n = s.size();
        vector<pair<int, int>> intervals(26, {-1, -1});
        for (int i = 0; i < n; ++i) {
            int c = s[i] - 'a';
            if (intervals[c].first == -1) {
                intervals[c].first = i;
            }
            intervals[c].second = i;
        }
        vector<pair<int, int>> valid_segments;
        for (int i = 0; i < n; ++i) {
            if (intervals[s[i] - 'a'].first == i) {
                int j = i + 1;
                int right = intervals[s[i] - 'a'].second;
                bool valid = true;
                while (j <= right) {
                    int c = s[j] - 'a';
                    if (intervals[c].first < i) {
                        valid = false;
                        break;
                    }
                    right = max(right, intervals[c].second);
                    ++j;
                }
                if (valid) {
                    valid_segments.emplace_back(i, j - 1);
                }
            }
        }
        vector<pair<int, int>> reduced_valid_segment;
        sort(valid_segments.begin(), valid_segments.end());
        reduced_valid_segment.emplace_back(valid_segments.back());
        int right = valid_segments.back().second;
        for (int i = (int)valid_segments.size() - 2; i >= 0; --i) {
            if (valid_segments[i].second < right) {
                reduced_valid_segment.emplace_back(valid_segments[i]);
                right = valid_segments[i].second;
            }
        }
        vector<string> ret;
        for (auto p : reduced_valid_segment) {
            ret.emplace_back(s.substr(p.first, p.second - p.first + 1));
        }
        return ret;
    }
};
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