<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 {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        // hash each string
        unordered_map<string, vector<string>> map;
        for (string s : strs) {
            vector<int> count(26, 0);
            for (char c : s) {
                count[c - 'a']++;
            }
            string hash = "";
            for(int i = 0; i < count.size(); i++) {
                if (count[i] != 0) {
                    hash += (i + 'a');
                    hash += to_string(count[i]);
                }
            }
            map[hash].push_back(s);
        }
        vector<vector<string>> result;
        for (auto p : map) {
            result.push_back(p.second);
        }
        return result;
    }
};
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