leetcode orb slam3 3/99--＞ leetcode49 Group Anagrams

Perfect — let’s break down all the knowledge and skills you need to confidently solve LeetCode 49: Group Anagrams. I’ll go step by step so you know exactly what to learn.

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1️⃣ Core Concepts

a. Strings

• Understand how to manipulate strings in your language of choice (Python, C++, Java).

• Know how to:

  • Sort a string: sorted(s) in Python, std::sort(s.begin(), s.end()) in C++.
  • Convert a string into a key/signature (like a tuple or string).

Example:

s = "eat"
key = "".join(sorted(s))  # "aet"

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b. Hashing / Maps

• A hash map (dictionary) is essential for grouping by keys.

• Know how to:

  • Insert items into a map: key → list of items.
  • Retrieve or update groups efficiently.

Example in Python:

groups = {}
for word in words:key = "".join(sorted(word))groups.setdefault(key, []).append(word)

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c. Arrays / Lists

• You need to store the groups in a list of lists or equivalent structure.
• Be comfortable iterating, appending, and returning lists.

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d. Sorting (optional variant)

• One common solution is to sort each string to generate a canonical form.
• Understand how sorting works in your language, and time complexity O(k log k) for a string of length k.

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2️⃣ Advanced/Optional Concepts

a. Counting / Frequency Array

• Instead of sorting, you can generate a character count signature:

count = [0] * 26
for c in word:count[ord(c) - ord('a')] += 1
key = tuple(count)

• This avoids sorting, faster for very long strings.

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b. Tuples / Immutable Keys

• Some languages (Python) require immutable types as keys in hash maps (tuple vs list).

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c. Complexity Analysis

• Understand time and space complexity:

  • Sorting each word: O(N*KlogK)
  • Counting letters: O(N*K)
  • Space: storing groups → O(N*K)

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✅ 3. Step-by-Step Knowledge Path

1. Basic string manipulation

   • Slicing, indexing, concatenation.

2. Hash maps / dictionaries

   • Key-value storage, insertion, retrieval, default values.

3. Sorting

   • Sorting strings or arrays.

4. Lists / arrays

   • Nested lists, append, iterate.

5. Optional: Counting / frequency arrays

   • Generate a signature without sorting.

6. Optional: Complexity reasoning

   • Compare sorting vs counting, analyze runtime.

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note1: explanation of anagrams

Ah! You want a clear explanation of anagrams itself. Let’s break it down simply.

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1️⃣ What is an Anagram?

An anagram is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

Examples:

• "listen" → "silent" ✅
• "eat" → "tea" ✅
• "rat" → "tar" ✅

Not an anagram:

• "hello" → "helo" ❌ (missing letter)
• "abc" → "abcd" ❌ (extra letter)

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2️⃣ Key Properties of Anagrams

1. Same letters
   Every letter in the original word must appear in the anagram.

2. Same counts
   If a letter appears twice in the original, it must appear twice in the anagram.

3. Order doesn’t matter
   "listen" and "silent" have different order but same letters → anagrams.

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3️⃣ How to Detect Anagrams

Method 1: Sort the letters

• Sort each word alphabetically.
• If two words have the same sorted form → they are anagrams.

Example:

"eat" -> "aet"
"tea" -> "aet"  # same → anagram

Method 2: Count letters

• Count frequency of each character.
• If counts match → they are anagrams.

Example:

"eat" -> {'e':1,'a':1,'t':1}
"tea" -> {'t':1,'e':1,'a':1}  # same counts → anagram

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4️⃣ Why it Matters in Programming

• Grouping problems: You can group all anagrams together (like LeetCode 49).
• Hashing: Sorting or counting gives a canonical form → makes grouping fast.
• Real-world analogy (like ORB-SLAM3): Different items that are “equivalent under some rule” (anagrams or feature descriptors) are mapped to the same key for efficient search.

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note2: canonical form

Ah! You want a clear explanation of canonical form in the context of LeetCode 49: Group Anagrams. Let’s break it down carefully.

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1️⃣ What is a Canonical Form?

A canonical form is a standardized representation of an object such that:

Two objects that are “equivalent” under some rules have exactly the same canonical form.

In Group Anagrams, the rule is:

Two strings are anagrams if they contain the same letters in the same counts.

So, the canonical form of a string is a representation that uniquely identifies all strings that are anagrams of each other.

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2️⃣ Examples

Strings: "eat", "tea", "ate"

Method 1: Sorted String

• Sort the letters alphabetically:

  "eat" → "aet"
  "tea" → "aet"
  "ate" → "aet"
  

• Result: "aet" is the canonical form for all three strings.

Method 2: Character Count (Frequency Array)

• Count letters a–z:

  "eat" → [1,0,0,...,1,...,1,...0] (1 for a, 1 for e, 1 for t)
  "tea" → [1,0,0,...,1,...,1,...0]
  "ate" → [1,0,0,...,1,...,1,...0]
  

• Tuple of counts [1,0,0,...,1,...,1,...0] → canonical form.

✅ Both methods produce a unique identifier for any group of anagrams.

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3️⃣ Why It’s Important

• Instead of comparing every pair of strings (O(n^2)), you hash the canonical form into a map.
• All anagrams map to the same key → easy to group in O(n * k) time (n = number of words, k = average length).

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4️⃣ ORB-SLAM3 Analogy

In ORB-SLAM3:

• Strings → ORB feature descriptors
• Canonical form → Visual word (quantized descriptor)
• Hash map → Inverted index for place recognition

Just like sorting letters or counting characters groups anagrams, ORB descriptors are quantized into visual words so that “similar descriptors” fall into the same bucket.

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note3: character count signature

Ah, got it! Let’s break down the “character count signature” in detail — this is a key alternative technique for LeetCode 49: Group Anagrams.

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1️⃣ What is a Character Count Signature?

Instead of sorting a string to get a canonical form, you can count how many times each character appears and use that as a signature (hashable key) for grouping.

Example

Input:

words = ["eat", "tea", "tan", "ate", "nat", "bat"]

• Count characters for "eat":

a:1, e:1, t:1

• Count characters for "tea":

a:1, e:1, t:1

Both counts are identical, so "eat" and "tea" belong to the same group.

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2️⃣ How to Represent It

You can store the counts as:

• Tuple of length 26 (for lowercase English letters) → immutable, hashable in Python.

count = [0] * 26
for c in word:count[ord(c) - ord('a')] += 1
key = tuple(count)

• String representation → join counts with separators (less efficient, but works).

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3️⃣ Why It Works

• Sorting: O(k log k) per word, where k = word length
• Counting: O(k) per word
• If the word is long, counting is faster.

Key idea: Different words that are anagrams produce the same count array, which acts as a signature/key for grouping.

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4️⃣ Mapping to ORB-SLAM3

• Words in LeetCode → ORB feature descriptors
• Count signature / sorted string → quantized descriptor or visual word
• Grouping by signature → inverted index in KeyFrameDatabase / Bag-of-Words grouping
• Both approaches hash items into buckets of similar items for fast retrieval.

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✅ Mini Example in Python

from collections import defaultdictwords = ["eat", "tea", "tan", "ate", "nat", "bat"]
groups = defaultdict(list)for word in words:count = [0]*26for c in word:count[ord(c)-ord('a')] += 1groups[tuple(count)].append(word)print(list(groups.values()))
# Output: [['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]

• tuple(count) is the character count signature — the “key” that groups anagrams.

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note4: groups = defaultdict(list)

Ah! Now I understand — you want a clear explanation of this line:

groups = defaultdict(list)

Let’s break it down step by step.

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1️⃣ What is defaultdict?

• defaultdict is a special type of dictionary from Python’s collections module.
• It works like a regular dictionary (dict) but has one important feature:

If you access a key that doesn’t exist yet, it automatically creates it with a default value you specify.

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2️⃣ How defaultdict(list) Works

• list is the default factory function.
• So when you do:

groups[some_key].append(some_value)

• If some_key is not yet in the dictionary, Python automatically does:

groups[some_key] = []

• Then it appends some_value to the list.

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Example

from collections import defaultdictgroups = defaultdict(list)groups['a'].append(1)  # 'a' doesn't exist yet, so a new list [] is created
groups['a'].append(2)
groups['b'].append(3)print(groups)

Output:

defaultdict(<class 'list'>, {'a': [1, 2], 'b': [3]})

✅ Notice that we didn’t have to check if 'a' or 'b' existed first. defaultdict did it automatically.

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3️⃣ Why it’s Useful in LeetCode 49

In Group Anagrams, you want to group words by a key (canonical form):

groups[key].append(word)

• key might not exist in the dictionary yet.
• defaultdict(list) ensures a new list is automatically created for that key, so you can safely append the word.

Without defaultdict, you’d have to do:

if key not in groups:groups[key] = []
groups[key].append(word)

defaultdict(list) makes your code cleaner and shorter.

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note5: groups[tuple(count)].append(word)

Ah! Now I understand — you want a step-by-step explanation of this line:

groups[tuple(count)].append(word)

Let’s break it down carefully.

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1️⃣ count

• count is a list of integers representing how many times each letter appears in the word.
• Example for "eat" (letters a–z):

count = [1,0,0, ..., 1, ..., 1,0]  # 1 for a, 1 for e, 1 for t, rest 0

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2️⃣ tuple(count)

• Lists in Python cannot be used as dictionary keys, because they are mutable.
• A tuple is immutable, so it can be used as a key.
• By converting the list to a tuple:

key = tuple(count)

• We now have a hashable key that uniquely represents the letter composition of the word.
• This key is identical for all anagrams of the word.

Example:

"eat" -> tuple(count) = (1,0,0,...,1,...,1,...,0)
"tea" -> tuple(count) = (1,0,0,...,1,...,1,...,0)  # same key

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3️⃣ groups[tuple(count)]

• groups is a defaultdict(list) — a dictionary where each key automatically has a list as a value.
• groups[tuple(count)] retrieves the list corresponding to that key.
• If the key doesn’t exist yet, a new empty list is created automatically.

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4️⃣ .append(word)

• We add the current word to the list corresponding to its canonical form (tuple of counts).
• All words with the same canonical form (i.e., anagrams) will be appended to the same list.

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🔹 Full Example

from collections import defaultdictwords = ["eat", "tea", "tan", "ate", "nat", "bat"]
groups = defaultdict(list)for word in words:count = [0]*26for c in word:count[ord(c)-ord('a')] += 1groups[tuple(count)].append(word)print(list(groups.values()))

Output:

[['eat', 'tea', 'ate'], ['tan', 'nat'], ['bat']]

• "eat", "tea", "ate" → same tuple → grouped together
• "tan", "nat" → same tuple → grouped together
• "bat" → unique tuple → its own group

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✅ Summary:

• count → frequency of letters in word
• tuple(count) → hashable key representing the “signature” of the word
• groups[tuple(count)] → retrieves the group of anagrams with the same signature
• .append(word) → adds the word to its group

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note6: immutable types

Ah! You want a clear explanation of immutable types and why they matter in Python. Let’s go step by step.

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1️⃣ What is an Immutable Type?

• Immutable types are objects whose value cannot be changed after they are created.
• Once you create an immutable object, you cannot modify it.

Examples of Immutable Types in Python:

• int → x = 5 (you can reassign x to 6, but the integer 5 itself cannot change)
• float → 3.14
• str → "hello"
• tuple → (1, 2, 3)

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2️⃣ What is a Mutable Type?

• Mutable types are objects that can be changed in place.
• You can modify their contents without creating a new object.

Examples:

• list → [1, 2, 3] → you can do append, pop, or change elements
• dict → {'a': 1} → you can add/remove keys
• set → {1, 2, 3} → you can add/remove elements

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3️⃣ Why Immutable Matters for Dictionary Keys

• Dictionary keys must be hashable.
• Hashable = the object has a hash value that does not change during its lifetime.
• Mutable objects like list are not hashable because you could change their content, which would break the dictionary.

d = {}
lst = [1,2,3]
d[lst] = "value"  # ❌ Error: unhashable type: 'list'

• Immutable objects like tuple are hashable → safe to use as dictionary keys.

d = {}
t = (1,2,3)
d[t] = "value"  # ✅ Works

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4️⃣ In LeetCode 49 (Group Anagrams)

• count is a list of letter frequencies → mutable → cannot be used as a dictionary key.
• Convert it to a tuple → immutable → can be used as a dictionary key:

groups[tuple(count)].append(word)

• Now all words with the same letter counts map to the same key, just like using a hashable signature.

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✅ Summary:

• Immutable = cannot change → hashable → can be dictionary key
• Mutable = can change → not hashable → cannot be dictionary key
• In Group Anagrams: tuple(count) is immutable, so it can serve as the canonical key for grouping words.

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note7 hashable

好的，我用中文来解释 hashable（可哈希） 的概念。

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1️⃣ 什么是可哈希对象（Hashable Object）？

可哈希对象 有两个特点：

1. 它有一个哈希值（hash(obj)），这个值唯一标识这个对象。
2. 它的哈希值在整个生命周期中不会改变。

Python 的字典（dict）就是基于哈希表实现的，所以 key 必须是可哈希的。

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2️⃣ 字典是如何工作的

当你执行：

d[key] = value

Python 会：

1. 计算 hash(key)。
2. 根据这个哈希值找到存储 value 的位置（槽位）。
3. 当你访问 d[key] 时，Python 重新计算 hash(key) 来找到 value。

✅ 因此 key 的哈希值必须稳定。

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3️⃣ 为什么可变对象不能作为 key

• 可变对象（如 list、dict）可以修改内容。

lst = [1, 2, 3]
d[lst] = "value"
lst.append(4)  # lst 改变了内容

• 如果 key 的内容改变，哈希值可能变化 → Python 找不到原来的 value
• 这会破坏字典的正常工作。

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4️⃣ 为什么元组（tuple）可以

• 元组是 不可变对象，内容不能改变
• 哈希值稳定，所以可以安全作为字典的 key：

t = (1,2,3)
d[t] = "value"  # 总是能取回

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5️⃣ 总结

属性	可变对象 (list)	不可变对象 (tuple)
内容可以改变？	✅ 可以	❌ 不可以
可哈希？	❌ 否	✅ 是
可以作为字典 key？	❌ 否	✅ 是

核心思想：

字典的 key 必须是可哈希的，这样 Python 才能计算出稳定的位置来存储和查找 value。

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note8: define count as a tuple?

Ah! You’re asking why we don’t define count as a tuple from the beginning instead of using a list and then converting it with tuple(count). Let me explain clearly.

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1️⃣ Lists vs Tuples in Python

• List ([]) → mutable

  • You can change elements: count[0] += 1

• Tuple (()) → immutable

  • Once created, elements cannot be changed

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2️⃣ Why count starts as a list

In LeetCode 49, we need to count the frequency of each character:

count = [0]*26
for c in word:count[ord(c)-ord('a')] += 1

• We modify each element of count as we iterate over the word.
• If we had defined count as a tuple:

count = (0, 0, 0, ..., 0)  # length 26
count[0] += 1  # ❌ Error! Tuples are immutable

• This would raise an error because tuples cannot be changed.

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3️⃣ Why we convert to a tuple at the end

• After counting is finished, we need a hashable key to use in the dictionary:

groups[tuple(count)].append(word)

• The tuple is immutable, so it can safely be used as a dictionary key.
• Converting at the end is efficient: we mutate the list while counting, then freeze it into a tuple for hashing.

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4️⃣ Summary

1. Count frequencies: need a mutable structure → list.
2. Use as dict key: need an immutable structure → convert to tuple.

✅ This is the standard Python trick: mutable while building, immutable for hashing.

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note9: the value of the dictionary (groups) is a list

Yes! In the line

groups[tuple(count)].append(word)

the value of the dictionary (groups) is a list. Let me explain clearly.

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1️⃣ What groups is

• groups is a defaultdict(list).

• This means:

  • The keys are hashable objects (here tuple(count)).
  • The values are lists by default.

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2️⃣ What .append(word) does

• groups[tuple(count)] retrieves the list associated with that key.
• .append(word) adds the current word to that list.

Example:

groups = defaultdict(list)groups[(1,0,0,...,1,...)]  # key = tuple(count)
groups[(1,0,0,...,1,...)] = []  # automatically created as empty list by defaultdict
groups[(1,0,0,...,1,...)].append("eat")  # add word to list

After adding more words with the same key:

groups[(1,0,0,...,1,...)] = ["eat", "tea", "ate"]

✅ So the value is a list of words that share the same canonical form.

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3️⃣ Summary

• Key: tuple(count) → represents the anagram signature.
• Value: a list → stores all words that match that signature.
• Why list? Because multiple words can be anagrams of each other, so we need to collect them together.

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note10: set is more efficient

Perfect! Here’s the more efficient version using a set instead of a list for each group. This makes checking if a word exists O(1) instead of O(n) in the list.

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1️⃣ Use defaultdict(set) instead of list

from collections import defaultdictwords = ["eat", "tea", "tan", "ate", "nat", "bat"]
groups = defaultdict(set)  # each value is a set instead of a listfor word in words:count = [0]*26for c in word:count[ord(c)-ord('a')] += 1groups[tuple(count)].add(word)  # add to set

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2️⃣ Function to check if a word exists

def word_exists(word, groups):count = [0]*26for c in word:count[ord(c)-ord('a')] += 1key = tuple(count)return word in groups.get(key, set())

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3️⃣ Example Usage

print(word_exists("tea", groups))  # True
print(word_exists("bat", groups))  # True
print(word_exists("cat", groups))  # False

• Now checking word in groups[key] is O(1) because sets are hash-based.
• This is especially useful if you have many words in each anagram group.

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✅ Summary of Improvements

Before	After
List of words	Set of words
word in list → O(n)	word in set → O(1)
Append: .append()	Add: .add()

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