First Non Repeating Character Using Hash in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of First Non Repeating Character Index

What is the output of the following Rust code that finds the index of the first non-repeating character in the string?

DSA Rust

use std::collections::HashMap;

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

fn main() {
    let s = "swiss";
    match first_non_repeating_char(s) {
        Some(i) => println!("{}", i),
        None => println!("-1"),
    }
}

A 2

B -1

C 4

D 1

💡 Hint

Count how many times each character appears, then find the first with count 1.

The string "swiss" has characters: s(3 times), w(1 time), i(1 time). The first non-repeating character is 'w' at index 1, so the output is 1.

❓ Predict Output

intermediate

Output of First Non Repeating Character Index with All Repeats

What is the output of the following Rust code when the input string has all repeating characters?

DSA Rust

use std::collections::HashMap;

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

fn main() {
    let s = "aabbcc";
    match first_non_repeating_char(s) {
        Some(i) => println!("{}", i),
        None => println!("-1"),
    }
}

A -1

B 0

C 5

D 6

💡 Hint

If no character appears once, the function returns None which prints -1.

All characters in "aabbcc" appear twice, so no non-repeating character exists. The function returns None, and the code prints -1.

🔧 Debug

advanced

Identify the Error in HashMap Usage

What error does the following Rust code produce when trying to find the first non-repeating character?

DSA Rust

use std::collections::HashMap;

fn first_non_repeating_char(s: &str) -> Option<usize> {
    let mut counts = HashMap::new();
    for ch in s.chars() {
        counts.insert(ch, counts[&ch] + 1);
    }
    for (i, ch) in s.chars().enumerate() {
        if counts[&ch] == 1 {
            return Some(i);
        }
    }
    None
}

fn main() {
    let s = "abc";
    match first_non_repeating_char(s) {
        Some(i) => println!("{}", i),
        None => println!("-1"),
    }
}

A Compilation error: cannot move out of borrowed content

B KeyError at counts[&ch]

C Panic at runtime: entry not found

D No error, outputs 0

💡 Hint

Check how counts[&ch] is accessed before insertion.

The code tries to access counts[&ch] before the key exists in the HashMap, causing a panic at runtime due to missing entry.

❓ Predict Output

advanced

Output of First Non Repeating Character Index with Unicode

What is the output of the following Rust code when the input string contains Unicode characters?

DSA Rust

use std::collections::HashMap;

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

fn main() {
    let s = "aáa";
    match first_non_repeating_char(s) {
        Some(i) => println!("{}", i),
        None => println!("-1"),
    }
}

A 2

B 1

C 0

D -1

💡 Hint

Unicode characters are treated as distinct chars.

The string "aáa" has 'a' twice and 'á' once at index 1. So the first non-repeating character is at index 1.

🧠 Conceptual

expert

Time Complexity of First Non Repeating Character Using HashMap

What is the time complexity of finding the first non-repeating character in a string of length n using a HashMap to count characters and then scanning again?

A O(n) because each character is processed twice but HashMap operations are O(1) average

B O(n^2) because nested loops are used to count characters

C O(log n) because HashMap uses balanced trees internally

D O(1) because HashMap lookup is constant time

💡 Hint

Consider how many times you iterate over the string and the cost of HashMap operations.

The algorithm iterates over the string twice: once to count characters and once to find the first unique character.

HashMap operations (insert, lookup) are average O(1), so total time is O(n).

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