<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:
    def firstUniqChar(self, s: str) -> int:
        alphabet_idx = [0 for i in range(26)]
        for ch in s:
            alphabet_idx[ord(ch) - ord("a")] += 1
        for i in range(len(s)):
            if alphabet_idx[ord(s[i])-ord("a")] == 1:
                return i
        return -1
        
            
        
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