First Non Repeating Character Using Hash in DSA Rust - Time & Space Complexity

Time Complexity: First Non Repeating Character Using Hash

O(n)

Understanding Time Complexity

We want to find how long it takes to find the first character that appears only once in a string using a hash map.

How does the time needed grow as the string gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

use std::collections::HashMap;

fn first_non_repeating_char(s: &str) -> Option {
    let mut counts = HashMap::new();
    for ch in s.chars() {
        *counts.entry(ch).or_insert(0) += 1;
    }
    for ch in s.chars() {
        if counts[&ch] == 1 {
            return Some(ch);
        }
    }
    None
}
    

This code counts how many times each character appears, then finds the first character with count one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops over the string characters.
• How many times: Each loop runs once over all characters, so twice in total.

How Execution Grows With Input

As the string gets longer, the number of steps grows roughly twice as fast because we scan the string two times.

Input Size (n)	Approx. Operations
10	About 20 steps
100	About 200 steps
1000	About 2000 steps

Pattern observation: The operations grow linearly with input size, doubling because of two passes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the first non-repeating character grows directly in proportion to the string length.

Common Mistake

[X] Wrong: "Because there are two loops, the time complexity is O(n²)."

[OK] Correct: Each loop runs separately over the string once, so their times add up, not multiply.

Interview Connect

Understanding how to analyze multiple passes over data is a key skill. This example shows how two simple scans still keep the process efficient.

Self-Check

What if we used nested loops to check each character against all others? How would the time complexity change?

Practice

1. What is the main purpose of using a hash map when finding the first non-repeating character in a string?

easy

A. To sort the characters alphabetically

B. To store the frequency of each character for quick lookup

C. To reverse the string efficiently

D. To convert characters to uppercase

Solution

1. Step 1: Understand the role of hash map

   A hash map stores each character as a key and its count as the value, allowing quick frequency lookup.

2. Step 2: Why frequency matters

   Knowing frequency helps identify which character appears only once, the first non-repeating character.

3. Final Answer:

   To store the frequency of each character for quick lookup -> Option B

4. Quick Check:

   Hash map = frequency storage [OK]

Hint: Use hash map to count characters fast [OK]

Common Mistakes:

• Thinking hash map sorts characters
• Using hash map to reverse string
• Ignoring frequency counts

2. Which Rust data structure is best suited to implement the frequency count for characters in a string?

easy

A. HashMap

B. Vec<char>

C. String

D. Option<char>

Solution

1. Step 1: Identify data structure for key-value pairs

   HashMap stores characters as keys and their counts as values, perfect for frequency counting.

2. Step 2: Why others don't fit

   Vec<char> is just a list, String holds characters but no counts, Option<char> holds one optional char only.

3. Final Answer:

   HashMap -> Option A

4. Quick Check:

   HashMap = char to count map [OK]

Hint: Use HashMap for frequency counts [OK]

Common Mistakes:

• Using Vec instead of HashMap
• Confusing String with frequency map
• Using Option for multiple counts

3. What will be the output of the following Rust code snippet?

use std::collections::HashMap;

fn first_non_repeating(s: &str) -> Option<char> {
    let mut counts = HashMap::new();
    for c in s.chars() {
        *counts.entry(c).or_insert(0) += 1;
    }
    for c in s.chars() {
        if counts[&c] == 1 {
            return Some(c);
        }
    }
    None
}

fn main() {
    let result = first_non_repeating("swiss");
    println!("{:?}", result);
}

medium

A. Some('s')

B. None

C. Some('i')

D. Some('w')

Solution

1. Step 1: Count character frequencies in "swiss"

   's' appears 3 times, 'w' once, 'i' once.

2. Step 2: Find first character with count 1 in original order

   Order: 's'(3), 'w'(1), 'i'(1), 's'(3), 's'(3). First with count 1 is 'w'.

3. Final Answer:

   Some('w') -> Option D

4. Quick Check:

   First unique char = 'w' [OK]

Hint: Check counts then first unique char in order [OK]

Common Mistakes:

• Picking first char without checking count
• Confusing 'i' and 'w' positions
• Ignoring multiple occurrences

4. Identify the bug in this Rust function to find the first non-repeating character:

fn first_non_repeating(s: &str) -> Option<char> {
    let mut counts = std::collections::HashMap::new();
    for c in s.chars() {
        counts.insert(c, 1);
    }
    for c in s.chars() {
        if counts[&c] == 1 {
            return Some(c);
        }
    }
    None
}

medium

A. Using insert overwrites counts, so frequencies are incorrect

B. Missing return type Option<char>

C. Using s.chars() instead of s.bytes()

D. No bug, code works correctly

Solution

1. Step 1: Analyze frequency counting loop

   Using counts.insert(c, 1) overwrites previous counts, so frequency is always 1.

2. Step 2: Effect on result

   All characters appear counted as 1, so function returns first char incorrectly.

3. Final Answer:

   Using insert overwrites counts, so frequencies are incorrect -> Option A

4. Quick Check:

   insert overwrites count, use entry().or_insert() [OK]

Hint: Use entry().or_insert() to update counts [OK]

Common Mistakes:

• Using insert instead of entry().or_insert()
• Ignoring frequency updates
• Confusing chars and bytes

5. Given a string that may contain Unicode characters, how would you modify the Rust function to correctly find the first non-repeating character considering Unicode graphemes instead of chars? Choose the best approach.

hard

A. Use chars() iterator as usual, it handles graphemes correctly

B. Convert string to bytes and count byte frequencies

C. Use the unicode-segmentation crate to iterate over graphemes and count frequencies

D. Replace all Unicode characters with ASCII equivalents before counting

Solution

1. Step 1: Understand Unicode graphemes vs chars

   Rust's chars() iterates Unicode scalar values, not grapheme clusters (user-perceived characters).

2. Step 2: Use unicode-segmentation crate

   This crate provides graphemes() iterator to correctly handle combined Unicode characters.

3. Step 3: Count frequencies using graphemes

   Counting graphemes ensures correct identification of non-repeating user-visible characters.

4. Final Answer:

   Use the unicode-segmentation crate to iterate over graphemes and count frequencies -> Option C

5. Quick Check:

   Unicode graphemes need special iterator [OK]

Hint: Use unicode-segmentation crate for grapheme counting [OK]

Common Mistakes:

• Assuming chars() equals graphemes
• Counting bytes instead of characters
• Replacing Unicode with ASCII loses data