<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 solve(string& s, int i, int j, vector<int>& dp) {
        if (i >= j) 
            return 0;
        if (dp[i * s.size() + j] != -1) 
            return dp[i * s.size() + j];
        if (s[i] == s[j]) 
            return dp[i * s.size() + j] = solve(s, i + 1, j - 1, dp);
        return dp[i * s.size() + j] = 1 + min(solve(s, i + 1, j, dp), solve(s, i, j - 1, dp));
    }

    int minInsertions(string s) {
        int n = s.size();
        vector<int> dp(n * n, -1);
        return solve(s, 0, n - 1, dp);  
    }
};

auto init = []() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();

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