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OJ notes/pages/Leetcode Group-Anagrams.md

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+# Leetcode Group-Anagrams
+
+2022-09-06 16:46
+
+> ##### Data structures:
+>
+> #DS #array #hash_table 
+>
+> ##### Difficulty:
+>
+> #coding_problem #difficulty-medium
+>
+> ##### Additional tags:
+>
+> #leetcode 
+>
+> ##### Revisions:
+>
+> N/A
+
+##### Links:
+
+- [Link to problem](https://leetcode.com/problems/group-anagrams/)
+
+---
+
+### Problem
+
+Given an array of strings `strs`, group **the anagrams** together. You can return the answer in **any order**.
+
+An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** strs = ["eat","tea","tan","ate","nat","bat"]
+**Output:** [["bat"],["nat","tan"],["ate","eat","tea"]]
+
+**Example 2:**
+
+**Input:** strs = [""]
+**Output:** [[""]]
+
+**Example 3:**
+
+**Input:** strs = ["a"]
+**Output:** [["a"]]
+
+#### Constraints
+
+-   `1 <= strs.length <= 104`
+-   `0 <= strs[i].length <= 100`
+-   `strs[i]` consists of lowercase English letters.
+
+### Thoughts
+
+> [!summary]
+> This is a #hash_table problem.
+
+This one can be solved simply using two-way hash tables, consisting two hash tables:
+- ana2id: turn anagram into an id
+- id2str: turn id into strings that has the anagrams
+
+The one problem is hashing vectors / arrays. Here we use
+CPP-style initialization of C's array.
+
+> [!tip]
+> CPP-style initialization of C array #tip
+> ```cpp
+> array<int, 26> ana = {};
+> ```
+
+Note that vectors doesn't work in unordered hash maps,
+unless we use custom hash functions.
+
+### Solution
+
+```cpp
+class Solution {
+public:
+  vector<vector<string>> groupAnagrams(vector<string> &strs) {
+    // using to hash maps for 2-way conversion
+    map<array<int, 26>, int> ana2id;
+    // this one is the answer returned.
+    vector<vector<string>> id2str;
+
+    int id = 0;
+    for (string str : strs) {
+      // Use aggregate initialization to default to 0.
+      array<int, 26> ana = {};
+
+      for (char s : str) {
+        ana[s - 'a']++;
+      }
+
+      if (ana2id.find(ana) == ana2id.end()) {
+        // Not in the map.
+        ana2id[ana] = id;
+        id2str.push_back({});
+        id2str.back().push_back(str);
+        id++;
+      } else {
+        // in the map, append it to id2str
+        id2str[ana2id[ana]].push_back(str);
+      }
+    }
+
+    return id2str;
+  }
+};
+```