String slices in Rust - Time & Space Complexity

Time Complexity: String slices

O(n)

Understanding Time Complexity

Let's see how the time it takes to work with string slices changes as the string gets bigger.

We want to know how fast operations on string slices run when the input grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn first_word(s: &str) -> &str {
    for (i, &item) in s.as_bytes().iter().enumerate() {
        if item == b' ' {
            return &s[0..i];
        }
    }
    &s[..]
}
    

This code finds the first word in a string slice by looking for the first space.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each byte of the string slice.
• How many times: Up to once for each byte until a space is found or the string ends.

How Execution Grows With Input

As the string gets longer, the loop may check more characters before finding a space or finishing.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The number of checks grows roughly in a straight line with the string length.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the first word grows directly with the length of the string slice.

Common Mistake

[X] Wrong: "Finding the first word always takes the same time no matter the string size."

[OK] Correct: If the first space is near the start, it is fast, but if the string has no spaces or the space is near the end, it takes longer because it checks more characters.

Interview Connect

Understanding how string slices work and their time cost helps you write efficient code and explain your choices clearly in real projects or interviews.

Self-Check

"What if we changed the code to find the last word instead of the first? How would the time complexity change?"

Practice

1. What is a string slice in Rust?

easy

A. A type of integer used for indexing strings

B. A reference to a part of a string without copying it

C. A function that modifies a string in place

D. A new string created by copying characters

Solution

1. Step 1: Understand string slice concept

   A string slice is a reference to a part of a string, not a new string.

2. Step 2: Compare options

   Only A reference to a part of a string without copying it describes a reference without copying, which matches string slices.

3. Final Answer:

   A reference to a part of a string without copying it -> Option B

4. Quick Check:

   String slice = reference without copy [OK]

Hint: Remember: slices borrow, not copy strings [OK]

Common Mistakes:

• Thinking slices create new strings
• Confusing slices with string modification
• Believing slices are integer types

2. Which of the following is the correct syntax to get a slice of the first 4 characters from a string text?

easy

A. &text[0..4]

B. text.slice(0, 4)

C. &text[1..5]

D. text[0:4]

Solution

1. Step 1: Recall Rust slice syntax

   Rust uses &string[start..end] to get a slice from start to end-1.

2. Step 2: Check options

   &text[0..4] uses &text[0..4], which correctly slices first 4 chars (indices 0 to 3).

3. Final Answer:

   &text[0..4] -> Option A

4. Quick Check:

   Slice syntax = &string[start..end] [OK]

Hint: Use &string[start..end] for slices, not methods or colons [OK]

Common Mistakes:

• Using method calls like slice() which don't exist
• Using 1-based indexing instead of 0-based
• Using Python-style slicing with colons

3. What will the following Rust code print?

let s = String::from("hello world");
let slice = &s[6..11];
println!("{}", slice);

medium

A. hello

B. Syntax error

C. hello world

D. world

Solution

1. Step 1: Understand slicing indices

   The slice &s[6..11] takes characters from index 6 up to 10 (11 excluded).

2. Step 2: Identify substring

   In "hello world", index 6 is 'w' and index 10 is 'd', so slice is "world".

3. Final Answer:

   world -> Option D

4. Quick Check:

   Slice &s[6..11] = "world" [OK]

Hint: Remember slice end index is excluded [OK]

Common Mistakes:

• Including the end index character
• Confusing start and end indices
• Expecting full string output

4. Identify the error in this Rust code snippet:

let s = String::from("rustacean");
let slice = &s[0..5];
println!("{}", slice);
let invalid_slice = &s[5..15];

medium

A. The slice &s[5..15] causes a runtime panic due to out-of-bounds

B. The slice &s[0..5] is invalid syntax

C. The println! macro is missing a semicolon

D. String::from cannot create strings with length 9

Solution

1. Step 1: Check slice ranges

   The string "rustacean" has length 9, so valid indices are 0 to 8.

2. Step 2: Analyze invalid slice

   The slice &s[5..15] tries to access indices beyond length, causing runtime panic.

3. Final Answer:

   The slice &s[5..15] causes a runtime panic due to out-of-bounds -> Option A

4. Quick Check:

   Out-of-bounds slice = panic [OK]

Hint: Slice end must be ≤ string length to avoid panic [OK]

Common Mistakes:

• Assuming slices auto-adjust to string length
• Thinking &s[0..5] is invalid syntax
• Ignoring runtime panics from invalid slices

5. Given a string text, how can you safely get a slice of the first n characters without risking a runtime panic if n is larger than the string length?

hard

A. Use text.get(0..n).unwrap()

B. Use &text[0..n] without checks

C. Use &text[0..n.min(text.len())]

D. Use &text[0..=n]

Solution

1. Step 1: Understand safe slicing

   To avoid panic, slice end must not exceed string length.

2. Step 2: Use min to limit slice end

   Using n.min(text.len()) ensures slice end is at most string length.

3. Final Answer:

   Use &text[0..n.min(text.len())] -> Option C

4. Quick Check:

   Safe slice end = min(n, length) [OK]

Hint: Use min() to limit slice end safely [OK]

Common Mistakes:

• Slicing without length check causing panic
• Using unwrap() which can panic if None
• Using inclusive range which is invalid for slices