<p>Given a string <code>s</code>, find the <strong>first</strong> non-repeating character in it and return its index. If it <strong>does not</strong> exist, return <code>-1</code>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;leetcode&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">0</span></p>

<p><strong>Explanation:</strong></p>

<p>The character <code>&#39;l&#39;</code> at index 0 is the first character that does not occur at any other index.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;loveleetcode&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;aabb&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">-1</span></p>
</div>

<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 only lowercase English letters.</li>
</ul>

class Solution {
public:
    int firstUniqChar(string s) {
        vector<int> freq(26, 0);
        queue<pair<char, int>> q;
        
        for (int i = 0; i < s.length(); i++) {
            char c = s[i];
            freq[c - 'a']++;

            if (freq[c - 'a'] == 1) {
                q.push({c, i});
            }
            
            while (!q.empty() && freq[q.front().first - 'a'] > 1) {
                q.pop();
            }
        }
        
        if (!q.empty()) {
            return q.front().second;
        }
        
        return -1;
    }
};
13700