<p>Given a string <code>s</code>. In one step you can insert any character at any index of the string.</p>

<p>Return <em>the minimum number of steps</em> to make <code>s</code>&nbsp;palindrome.</p>

<p>A&nbsp;<b>Palindrome String</b>&nbsp;is one that reads the same backward as well as forward.</p>

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

<pre>
<strong>Input:</strong> s = &quot;zzazz&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> The string &quot;zzazz&quot; is already palindrome we do not need any insertions.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;mbadm&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> String can be &quot;mbdadbm&quot; or &quot;mdbabdm&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> Inserting 5 characters the string becomes &quot;leetcodocteel&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
    int lcs(string& s, string& t) {
	    int n = (int)s.size(), m = (int)t.size();
	    vector<int> prev(m + 1), cur;
	    for (int i = n - 1; i >= 0; i--) {
		    cur.assign(m + 1, 0);
		    for (int j = m - 1; j >= 0; j--) {
		    	if (s[i] == t[j])
			    	cur[j] = prev[j + 1] + 1;
			    else
			    	cur[j] = max(prev[j], cur[j + 1]);
		    }
		    prev = cur;
    	}
	    return prev[0];
    }
public:
    int minInsertions(string s) {
        string t = s;
        reverse(t.begin(), t.end());
        return (int)s.size() - lcs(s, t);
    }
};
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