<p>You are given an array of <code>n</code> strings <code>strs</code>, all of the same length.</p>

<p>We may choose any deletion indices, and we delete all the characters in those indices for each string.</p>

<p>For example, if we have <code>strs = [&quot;abcdef&quot;,&quot;uvwxyz&quot;]</code> and deletion indices <code>{0, 2, 3}</code>, then the final array after deletions is <code>[&quot;bef&quot;, &quot;vyz&quot;]</code>.</p>

<p>Suppose we chose a set of deletion indices <code>answer</code> such that after deletions, the final array has <strong>every string (row) in lexicographic</strong> order. (i.e., <code>(strs[0][0] &lt;= strs[0][1] &lt;= ... &lt;= strs[0][strs[0].length - 1])</code>, and <code>(strs[1][0] &lt;= strs[1][1] &lt;= ... &lt;= strs[1][strs[1].length - 1])</code>, and so on). Return <em>the minimum possible value of</em> <code>answer.length</code>.</p>

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

<pre>
<strong>Input:</strong> strs = [&quot;babca&quot;,&quot;bbazb&quot;]
<strong>Output:</strong> 3
<strong>Explanation:</strong> After deleting columns 0, 1, and 4, the final array is strs = [&quot;bc&quot;, &quot;az&quot;].
Both these rows are individually in lexicographic order (ie. strs[0][0] &lt;= strs[0][1] and strs[1][0] &lt;= strs[1][1]).
Note that strs[0] &gt; strs[1] - the array strs is not necessarily in lexicographic order.</pre>

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

<pre>
<strong>Input:</strong> strs = [&quot;edcba&quot;]
<strong>Output:</strong> 4
<strong>Explanation:</strong> If we delete less than 4 columns, the only row will not be lexicographically sorted.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> strs = [&quot;ghi&quot;,&quot;def&quot;,&quot;abc&quot;]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All rows are already lexicographically sorted.
</pre>

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

<ul>
	<li><code>n == strs.length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= strs[i].length &lt;= 100</code></li>
	<li><code>strs[i]</code> consists of lowercase English letters.</li>
</ul>

<ul>
	<li>&nbsp;</li>
</ul>

class Solution {
public:
    vector<vector<int>> gra;
    vector<int> depth;
    // return max dep starting from src
    int dfs(int src)
    {
        if(depth[src]!=-1) return depth[src];
        int maxm = 0;
        for(auto& ne : gra[src])
        {
            maxm = max(maxm, dfs(ne));
        }
        return depth[src]=maxm+1;
    }
    int minDeletionSize(vector<string>& s) 
    {
        int n = s.size();
        int m = s[0].size();
        gra.resize(m, vector<int>());
        depth.resize(m, -1);
        for(int j1=0; j1<m; j1++)
        {
            for(int j2 = j1+1; j2 < m; j2++)
            {
                bool ordered = true;
                for(int i=0; i<n; i++)
                {
                    if(s[i][j1]>s[i][j2]) {
                        ordered = false;
                        break;
                    }
                }
                if(ordered) 
                    gra[j1].push_back(j2);
            }
        }
        for(int i= 0; i<m; i++)
        {
            dfs(i);
        }
        int maxm = 0;
        for(int i= 0; i<m; i++)
        {
            maxm = max(maxm, depth[i]);
        }
        return m-maxm;
    }
};
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