<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 f(string s, string t) {
    int n = s.size();
    int m = t.size();
    
    vector<int> prev(m+1, 0), cur(m+1, 0);  // Only two 1D arrays for space optimization
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (s[i-1] == t[j-1]) {
                cur[j] = 1 + prev[j-1];
            } else {
                cur[j] = max(prev[j], cur[j-1]);
            }
        }
        prev = cur;  // Move current row to previous row for the next iteration
    }
    
    return prev[m];
}
int lcs(string s, string t){
    int n=s.size();
    int m=t.size();
    vector<int> cur(m+1);
    vector<int> prev(m+1);
    return f(s,t);
}
 
    int minInsertions(string s) {
        int size=s.length();
        string n= s;
        reverse(n.begin(),n.end());
        int staticLen= lcs(s,n);
        return size - staticLen; 
    }
};
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