<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 {
public:
    int LCS(string X, string Y){
        int m=X.length();
        int t[1001][1001] = {0};
        for(int i=1;i<=m;i++){
            for(int j=1;j<=m;j++){
                if(X[i-1] == Y[j-1]) t[i][j] = 1+ t[i-1][j-1];
                else t[i][j] = max(t[i-1][j] , t[i][j-1]);
            }
        }
        return t[m][m];
    }

    int minInsertions(string s) {
        string s1,s2;
        // doing 2 times reversal so that original string remains same
        s1 = s;
        reverse(s1.begin(),s1.end());
        s2 = s1; // reversed string
        reverse(s1.begin(),s1.end());

        return s.length() - LCS(s1,s2);
    }
};
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