Longest Common Prefix in DSA Rust - Build from Scratch

Longest Common Prefix

📖 Scenario: You work in a company that manages a list of product codes. You want to find the longest common starting part (prefix) shared by all product codes to help organize them better.

🎯 Goal: Build a Rust program that finds the longest common prefix among a list of product codes.

📋 What You'll Learn

Create a vector of strings called product_codes with exact values

Create a variable called prefix to store the longest common prefix

Write a function called longest_common_prefix that takes a vector of strings and returns a string

Print the longest common prefix stored in prefix

💡 Why This Matters

🌍 Real World

Finding common prefixes helps in organizing product codes, file paths, or DNA sequences in real applications.

💼 Career

Understanding string processing and writing clean functions is essential for software development and data processing jobs.

1

Create the list of product codes

Create a vector of strings called product_codes with these exact values: "flower", "flow", "flight"

DSA Rust

# Create the vector product_codes with the given strings
# Your code here

Hint

Use vec!["string1".to_string(), "string2".to_string(), ...] to create a vector of strings.

2

Create a variable to hold the prefix

Create a mutable variable called prefix and set it to an empty string ""

DSA Rust

fn main() {
    let product_codes = vec!["flower".to_string(), "flow".to_string(), "flight".to_string()];
    // Create mutable variable prefix as empty string
    # Your code here
}

Hint

Use let mut prefix = String::new(); to create an empty mutable string.

3

Write the function to find the longest common prefix

Write a function called longest_common_prefix that takes a vector of strings called strs and returns a String. Implement the logic to find the longest common prefix among all strings in strs. Then call this function with product_codes and assign the result to prefix.

DSA Rust

fn main() {
    let product_codes = vec!["flower".to_string(), "flow".to_string(), "flight".to_string()];
    let mut prefix = String::new();
    // Write the function longest_common_prefix and assign its result to prefix
    # Your code here
}

Hint

Start with the first string as prefix. For each other string, reduce prefix until it matches the start of that string.

4

Print the longest common prefix

Write a println! statement to print the value of prefix

DSA Rust

fn longest_common_prefix(strs: Vec) -> String {
    if strs.is_empty() {
        return String::new();
    }
    let mut prefix = strs[0].clone();
    for s in strs.iter().skip(1) {
        while !s.starts_with(&prefix) {
            prefix.pop();
            if prefix.is_empty() {
                return String::new();
            }
        }
    }
    prefix
}

fn main() {
    let product_codes = vec!["flower".to_string(), "flow".to_string(), "flight".to_string()];
    let mut prefix = String::new();
    prefix = longest_common_prefix(product_codes);
    // Print the prefix
    # Your code here
}

Hint

Use println!("{}", prefix); to print the prefix.

Practice

1. What does the Longest Common Prefix (LCP) of a list of strings represent?

easy

A. The shortest string in the list

B. The longest substring found anywhere in all strings

C. The longest starting substring common to all strings

D. The last character of the longest string

Solution

1. Step 1: Understand the prefix concept

   A prefix is a substring starting at the beginning of a string.

2. Step 2: Define longest common prefix

   The longest common prefix is the longest substring starting at the beginning that all strings share.

3. Final Answer:

   The longest starting substring common to all strings -> Option C

4. Quick Check:

   Prefix means start of strings = A [OK]

Hint: Focus on start of strings, not anywhere inside [OK]

Common Mistakes:

• Confusing prefix with substring anywhere
• Choosing shortest string instead of common start
• Thinking prefix is the last part of strings

2. Which Rust code snippet correctly initializes a vector of strings for finding the longest common prefix?

easy

A. let words = vec!["flower", "flow", "flight"];

B. let words = ["flower", "flow", "flight"];

C. let words = vec!("flower", "flow", "flight");

D. let words = vec<str>["flower", "flow", "flight"];

Solution

1. Step 1: Identify correct vector initialization syntax

   In Rust, vec![...] creates a vector with elements inside square brackets.

2. Step 2: Check syntaxes

   vec![...] is correct. [...] without vec! creates an array, not a vector. vec!() uses incorrect parentheses. vec<str>[...] is invalid syntax.

3. Final Answer:

   let words = vec!["flower", "flow", "flight"]; -> Option A

4. Quick Check:

   vec![...] is correct vector syntax [OK]

Hint: Use vec![...] to create vectors in Rust [OK]

Common Mistakes:

• Using parentheses instead of square brackets in vec!
• Confusing arrays with vectors
• Incorrect generic syntax for vec

3. What is the output of this Rust code that finds the longest common prefix?

fn longest_common_prefix(strs: Vec<&str>) -> String {
    if strs.is_empty() { return String::new(); }
    let mut prefix = strs[0].to_string();
    for s in &strs[1..] {
        while !s.starts_with(&prefix) {
            prefix.pop();
            if prefix.is_empty() { return String::new(); }
        }
    }
    prefix
}

fn main() {
    let words = vec!["interview", "internet", "internal"];
    println!("{}", longest_common_prefix(words));
}

medium

A. "int"

B. "inter"

C. "in"

D. "internal"

Solution

1. Step 1: Initialize prefix with first string

   Prefix starts as "interview".

2. Step 2: Compare prefix with second string "internet"

   "internet".starts_with("interview") is false, so pop last char repeatedly: pop 'w' -> "intervie" (false), pop 'e' -> "intervi" (false), pop 'i' -> "interv" (false), pop 'v' -> "inter" (true, since "internet".starts_with("inter")).

3. Step 3: Compare prefix "inter" with third string "internal"

   "internal".starts_with("inter") is true, so prefix remains "inter".

4. Final Answer:

   "inter" -> Option B

5. Quick Check:

   Common start is "inter" = B [OK]

Hint: Shrink prefix until all strings start with it [OK]

Common Mistakes:

• Not shrinking prefix correctly
• Assuming full first string is prefix
• Confusing substring anywhere with prefix

4. Identify the bug in this Rust function for longest common prefix and select the fix:

fn longest_common_prefix(strs: Vec<&str>) -> String {
    if strs.is_empty() { return String::new(); }
    let mut prefix = strs[0].to_string();
    for s in &strs[1..] {
        while !s.starts_with(prefix) {
            prefix.pop();
            if prefix.is_empty() { return String::new(); }
        }
    }
    prefix
}

medium

A. Change s.starts_with(prefix) to s.starts_with(&prefix)

B. Replace prefix.pop() with prefix.remove(0)

C. Initialize prefix as empty string instead of strs[0]

D. Add a break statement inside the while loop

Solution

1. Step 1: Identify type mismatch in starts_with

   starts_with expects a &str reference, but prefix is String. Passing prefix directly causes type error.

2. Step 2: Fix by passing reference to prefix

   Use &prefix to pass a &str reference, matching function signature.

3. Final Answer:

   Change s.starts_with(prefix) to s.starts_with(&prefix) -> Option A

4. Quick Check:

   Use & prefix for &str parameter [OK]

Hint: Pass &String as &str in starts_with [OK]

Common Mistakes:

• Passing String instead of &str to starts_with
• Removing first char instead of last char
• Not checking empty prefix before popping

5. Given a list of strings, write a Rust function that returns the longest common prefix but ignores case differences (case-insensitive). Which approach is best?

hard

A. Reverse strings and find prefix from the end

B. Compare characters using ASCII values directly

C. Use starts_with without modification

D. Convert all strings to lowercase before finding prefix

Solution

1. Step 1: Understand case-insensitive comparison

   To ignore case, convert all strings to the same case (lowercase or uppercase) before comparison.

2. Step 2: Apply lowercase conversion before prefix logic

   Transform each string to lowercase, then find longest common prefix normally.

3. Final Answer:

   Convert all strings to lowercase before finding prefix -> Option D

4. Quick Check:

   Normalize case first for case-insensitive prefix [OK]

Hint: Normalize case before prefix check [OK]

Common Mistakes:

• Comparing characters without case normalization
• Using ASCII values without case mapping
• Reversing strings which doesn't solve case issue