<p>Given an integer <code>n</code>, return <em>a list of all <strong>simplified</strong> fractions between </em><code>0</code><em> and </em><code>1</code><em> (exclusive) such that the denominator is less-than-or-equal-to </em><code>n</code>. You can return the answer in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> [&quot;1/2&quot;]
<strong>Explanation:</strong> &quot;1/2&quot; is the only unique fraction with a denominator less-than-or-equal-to 2.
</pre>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> [&quot;1/2&quot;,&quot;1/3&quot;,&quot;2/3&quot;]
</pre>

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

<pre>
<strong>Input:</strong> n = 4
<strong>Output:</strong> [&quot;1/2&quot;,&quot;1/3&quot;,&quot;1/4&quot;,&quot;2/3&quot;,&quot;3/4&quot;]
<strong>Explanation:</strong> &quot;2/4&quot; is not a simplified fraction because it can be simplified to &quot;1/2&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
</ul>

class Solution:
    def gcd(self, a, b):
        if b == 0:
            return a
        return self.gcd(b, a % b)
    
    def simplifiedFractions(self, n: int) -> list[str]:
        fractions = []
        for i in range(1,n+1):
            for j in range(1,i):
                if self.gcd(i,j) == 1:
                    fractions.append(f"{j}/{i}")
        return fractions
17100