<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 helper(int left, int right, const string &s, vector<vector<int>> &dp) {
        if (left >= right) return 0;  // No insertions needed when indices cross or single character

        if (dp[left][right] != -1) return dp[left][right];

        if (s[left] == s[right]) {
            // No insertion needed if characters match
            return dp[left][right] = helper(left + 1, right - 1, s, dp);
        }

        // Insert one character either on the left or the right
        return dp[left][right] = min(1 + helper(left + 1, right, s, dp), 
                                     1 + helper(left, right - 1, s, dp));
    }

public:
    int minInsertions(string s) {
        int n = s.size();
        vector<vector<int>> dp(n, vector<int>(n, -1));  // Memoization table
        return helper(0, n - 1, s, dp);  // Start from the full range of the string
    }
};

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