<p>Given an array of strings <code>strs</code>, group the <span data-keyword="anagram">anagrams</span> together. You can return the answer in <strong>any order</strong>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">strs = [&quot;eat&quot;,&quot;tea&quot;,&quot;tan&quot;,&quot;ate&quot;,&quot;nat&quot;,&quot;bat&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;bat&quot;],[&quot;nat&quot;,&quot;tan&quot;],[&quot;ate&quot;,&quot;eat&quot;,&quot;tea&quot;]]</span></p>

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

<ul>
	<li>There is no string in strs that can be rearranged to form <code>&quot;bat&quot;</code>.</li>
	<li>The strings <code>&quot;nat&quot;</code> and <code>&quot;tan&quot;</code> are anagrams as they can be rearranged to form each other.</li>
	<li>The strings <code>&quot;ate&quot;</code>, <code>&quot;eat&quot;</code>, and <code>&quot;tea&quot;</code> are anagrams as they can be rearranged to form each other.</li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">strs = [&quot;&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;&quot;]]</span></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">strs = [&quot;a&quot;]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;a&quot;]]</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= strs.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= strs[i].length &lt;= 100</code></li>
	<li><code>strs[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        checked = []
        result = []
        n = len(strs)

        for i in range(n):

            if (strs[i] in checked):
                continue
            s = strs[i]
            curr = []
            curr.append(s)

            for j in range(i+1,n):
                t = strs[j]

                if (isAnagrams(s,t)):
                    curr.append(t)
                    checked.append(t)
                     
            result.append(curr)

        return result

def isAnagrams(s, t):
    if (len(s) != len(t)):
        return False

    dict_s = {}
    dict_t = {}

    for letter in s:
        if (letter not in dict_s):
            dict_s[letter] = 1
        else:
            dict_s[letter] += 1
    
    for letter in t:
        if (letter not in dict_t):
            dict_t[letter] = 1
        else:
            dict_t[letter] += 1
    
    for key, value in dict_s.items():
        if (key not in dict_t):
            return False
        if (value != dict_t[key]):
            return False
    
    return True      
18400