Group Anagrams Using Hash Map in DSA Rust - Time & Space Complexity

Time Complexity: Group Anagrams Using Hash Map

O(n * k log k)

Understanding Time Complexity

We want to understand how long it takes to group words that are anagrams using a hash map.

How does the time grow when we have more words or longer words?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

use std::collections::HashMap;

fn group_anagrams(strs: Vec) -> Vec> {
    let mut map: HashMap> = HashMap::new();
    for s in strs {
        let mut chars: Vec = s.chars().collect();
        chars.sort_unstable();
        let key: String = chars.into_iter().collect();
        map.entry(key).or_default().push(s);
    }
    map.into_values().collect()
}
    

This code groups words that are anagrams by sorting each word and using the sorted word as a key in a hash map.

Identify Repeating Operations

• Primary operation: Loop over all words and sort each word's characters.
• How many times: Once for each word in the input list.

How Execution Grows With Input

As the number of words grows, the code sorts each word, so the time depends on both the number of words and their length.

Input Size (n)	Approx. Operations
10 words, length 5	About 10 x 5 log 5 ≈ 10 x 11.6 = 116 operations
100 words, length 5	About 100 x 5 log 5 ≈ 100 x 11.6 = 1160 operations
1000 words, length 5	About 1000 x 5 log 5 ≈ 1000 x 11.6 = 11600 operations

Pattern observation: The time grows roughly linearly with the number of words and also depends on the length of each word times its log.

Final Time Complexity

Time Complexity: O(n * k log k)

This means the time grows with the number of words (n) and the time to sort each word of length k.

Common Mistake

[X] Wrong: "Sorting each word is constant time, so the whole process is just O(n)."

[OK] Correct: Sorting depends on word length, so longer words take more time. Ignoring this misses an important part of the cost.

Interview Connect

Understanding how sorting each word affects the total time helps you explain your solution clearly and shows you know what matters when handling many or long words.

Self-Check

What if we used a character count array as the key instead of sorting? How would the time complexity change?

Practice

1. What is the main idea behind grouping anagrams using a hash map?

easy

A. Use the length of words as keys to group them

B. Store words in a list without any grouping

C. Group words by their first letter only

D. Use sorted strings as keys to group words with the same letters

Solution

1. Step 1: Understand anagram grouping

   Anagrams have the same letters in different orders, so sorting letters gives a common key.

2. Step 2: Use sorted string as hash map key

   By sorting each word and using it as a key, all anagrams map to the same key and group together.

3. Final Answer:

   Use sorted strings as keys to group words with the same letters -> Option D

4. Quick Check:

   Sorted string key groups anagrams [OK]

Hint: Sort each word to find its anagram group key [OK]

Common Mistakes:

• Grouping by word length only
• Using first letter as key
• Not sorting letters before grouping

2. Which Rust code snippet correctly creates a hash map to group anagrams by sorted keys?

easy

A. let mut map: HashMap = HashMap::new();

B. let map = Vec::new();

C. let mut map: HashMap = HashMap::new();

D. let mut map: HashMap = HashMap::new();

Solution

1. Step 1: Identify correct hash map type

   We need keys as sorted strings and values as vectors of strings to hold grouped anagrams.

2. Step 2: Match Rust syntax for hash map

   HashMap correctly maps sorted keys to lists of words.

3. Final Answer:

   let mut map: HashMap = HashMap::new(); -> Option A

4. Quick Check:

   HashMap with String keys and Vec<String> values [OK]

Hint: Use HashMap for grouping anagrams [OK]

Common Mistakes:

• Using Vec instead of HashMap
• Wrong value type (String instead of Vec<String>)
• Incorrect key type (i32 instead of String)

3. What is the output of this Rust code that groups anagrams?

use std::collections::HashMap;

fn group_anagrams(words: Vec<&str>) -> Vec {
    let mut map: HashMap = HashMap::new();
    for word in words {
        let mut chars: Vec<char> = word.chars().collect();
        chars.sort_unstable();
        let key = chars.into_iter().collect::<String>();
        map.entry(key).or_default().push(word.to_string());
    }
    map.into_values().collect()
}

fn main() {
    let input = vec!["eat", "tea", "tan", "ate", "nat", "bat"];
    let result = group_anagrams(input);
    println!("{:?}", result);
}

medium

A. [["eat", "ate", "tea"], ["tan", "nat"], ["bat"]]

B. [["eat"], ["tea"], ["tan"], ["ate"], ["nat"], ["bat"]]

C. [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]]

D. [["eat", "tea"], ["tan"], ["ate", "nat"], ["bat"]]

Solution

1. Step 1: Sort each word to create keys

   "eat", "tea", "ate" all sort to "aet"; "tan", "nat" to "ant"; "bat" to "abt".

2. Step 2: Group words by sorted keys

   Words with same sorted key are grouped: ["eat", "tea", "ate"], ["tan", "nat"], ["bat"].

3. Final Answer:

   [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]] -> Option C

4. Quick Check:

   Sorted keys group anagrams [OK]

Hint: Sort letters and group words by sorted key [OK]

Common Mistakes:

• Not sorting letters before grouping
• Grouping single words separately
• Mixing order of groups incorrectly

4. Identify the bug in this Rust code for grouping anagrams:

use std::collections::HashMap;

fn group_anagrams(words: Vec<&str>) -> Vec {
    let mut map: HashMap = HashMap::new();
    for word in words {
        let mut chars: Vec<char> = word.chars().collect();
        chars.sort(); // Bug here
        let key = chars.into_iter().collect::<String>();
        map.entry(key).or_default().push(word.to_string());
    }
    map.into_values().collect()
}

medium

A. Use sort_unstable() instead of sort() for better performance

B. Change chars.into_iter() to chars.iter()

C. Use a HashSet instead of HashMap

D. Convert word to lowercase before sorting

Solution

1. Step 1: Identify sorting method issue

   Using chars.sort() works but sort_unstable() is preferred for primitive types like char for speed.

2. Step 2: Explain why sort_unstable() is better

   sort_unstable() is faster and sufficient here because order stability is not needed for grouping anagrams.

3. Final Answer:

   Use sort_unstable() instead of sort() for better performance -> Option A

4. Quick Check:

   sort_unstable() improves sorting speed [OK]

Hint: Use sort_unstable() for sorting chars in anagrams [OK]

Common Mistakes:

• Using sort() unnecessarily slowing code
• Changing iterator type incorrectly
• Ignoring case sensitivity issues

5. Given a list of words with mixed case and spaces, how would you modify the Rust code to group anagrams correctly ignoring case and spaces? Example input: ["Listen", "Silent", "en list", "in lets"]

hard

A. Sort letters without any changes to words

B. Convert words to lowercase and remove spaces before sorting letters

C. Use word length as key for grouping

D. Group words by their first character only

Solution

1. Step 1: Normalize words by removing spaces and converting to lowercase

   To group anagrams ignoring case and spaces, clean words by removing spaces and making all letters lowercase.

2. Step 2: Sort letters of normalized words and use as keys

   After normalization, sort letters to get keys that group all anagrams regardless of case or spaces.

3. Final Answer:

   Convert words to lowercase and remove spaces before sorting letters -> Option B

4. Quick Check:

   Normalize words before sorting for correct grouping [OK]

Hint: Lowercase and remove spaces before sorting letters [OK]

Common Mistakes:

• Not normalizing case and spaces
• Using length or first letter as key
• Sorting original words without cleaning