<p>Given a string <code>s</code>, return <code>true</code> <em>if the </em><code>s</code><em> can be palindrome after deleting <strong>at most one</strong> character from it</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abca&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> You could delete the character &#39;c&#39;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
   bool validPalindrome(string s) {
        unordered_map<char, int> charCount;
        
        // Count the frequency of each character
        for (char a : s) {
            charCount[a]++;
        }

        // Count the number of characters with odd frequencies
        int oddCount = 0;
        for (auto x : charCount) {
            if (x.second % 2 != 0) {
                oddCount++;
            }
        }

        // A string can be a palindrome if:
        // - It has at most one character with an odd frequency
        // However, this is not directly relevant to removing one character.
        // We need to use a two-pointer approach.
        
        int left = 0, right = s.size() - 1;
        while (left < right) {
            if (s[left] != s[right]) {
                // Try skipping either the left character or the right character
                return isPalindrome(s, left + 1, right) || isPalindrome(s, left, right - 1);
            }
            left++;
            right--;
        }
        
        // If we traverse without mismatches, it's a valid palindrome
        return true;
    }

    // Helper function to check if a substring is a palindrome
    bool isPalindrome(const string& s, int left, int right) {
        while (left < right) {
            if (s[left] != s[right]) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
};
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