Longest Common Prefix in DSA Rust - Time & Space Complexity

Time Complexity: Longest Common Prefix

O(n * m)

Understanding Time Complexity

We want to understand how the time needed to find the longest common prefix changes as we add more words or longer words.

How does the work grow when input size grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn longest_common_prefix(strs: &[String]) -> 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
}

This code finds the longest common prefix string among an array of strings by shrinking the prefix until all strings start with it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking if each string starts with the current prefix and shortening the prefix if not.
• How many times: For each of the n strings, the prefix may be shortened up to m times, where m is the length of the first string.

How Execution Grows With Input

As the number of strings (n) or the length of the prefix (m) grows, the number of checks grows roughly by n times m.

Input Size (n)	Approx. Operations
10 strings, prefix length 5	About 50 checks
100 strings, prefix length 10	About 1000 checks
1000 strings, prefix length 20	About 20,000 checks

Pattern observation: The work grows roughly by multiplying the number of strings by the prefix length.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally to the number of strings times the length of the prefix.

Common Mistake

[X] Wrong: "The time depends only on the number of strings, not their length."

[OK] Correct: The prefix length affects how many times we shorten and check strings, so longer prefixes mean more work.

Interview Connect

Understanding this helps you explain how string operations affect performance and shows you can analyze nested loops clearly.

Self-Check

"What if we sorted the strings first and only compared the first and last strings? How would the time complexity change?"

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