<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 s1, string s2) {
        int n=s1.size();
        int m=s2.size();

        // vector<vector<int>>  dp(n+1,vector<int>(m+1,0));
        vector<int> prev(m+1,0);
        vector<int> cur(m+1,0);


        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(s1[i-1]==s2[j-1]){
                    cur[j]=1+prev[j-1];

                }
                else cur[j] = max(prev[j] , cur[j-1]);
            }
            prev=cur;
            
        }
        return prev[m];
    }

    int lps(string s) {
        stack <int> st;
        string t="";
        for(auto it:s){
            st.push(it);

        }

        while(!st.empty()){
            t+=st.top();
            st.pop();
        }

        return lcs(s,t);
        
    }
    int minInsertions(string s) {
        return s.size()-lps(s);
    }
};
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