First and Last Occurrence of Element in DSA Rust - Build from Scratch

First and Last Occurrence of Element

📖 Scenario: You are working with a list of numbers representing daily temperatures recorded over a week. You want to find the first and last day a specific temperature was recorded.

🎯 Goal: Build a Rust program that finds the first and last position of a given temperature in a list.

📋 What You'll Learn

Create a vector called temperatures with exact values: [23, 25, 23, 22, 25, 23, 24]

Create a variable called target and set it to 23

Use a for loop with index i to find the first and last occurrence of target in temperatures

Print the first and last occurrence indices in the format: First occurrence: X and Last occurrence: Y

💡 Why This Matters

🌍 Real World

Finding first and last occurrences is useful in analyzing time series data, logs, or any ordered data to detect events or patterns.

💼 Career

This skill helps in debugging, data analysis, and building efficient search features in software development.

1

Create the temperature list

Create a vector called temperatures with these exact values: [23, 25, 23, 22, 25, 23, 24]

DSA Rust

# Create the vector temperatures with the given values
# Your code here

Hint

Use let temperatures = vec![23, 25, 23, 22, 25, 23, 24]; to create the vector.

2

Set the target temperature

Create a variable called target and set it to 23

DSA Rust

fn main() {
    let temperatures = vec![23, 25, 23, 22, 25, 23, 24];
    // Create target variable
    // Your code here
}

Hint

Use let target = 23; to set the target temperature.

3

Find first and last occurrence

Use a for loop with index i to find the first and last occurrence of target in temperatures. Create variables first and last to store these positions. Initialize first to None and last to None. Update them inside the loop.

DSA Rust

fn main() {
    let temperatures = vec![23, 25, 23, 22, 25, 23, 24];
    let target = 23;
    let mut first: Option = None;
    let mut last: Option = None;
    // Use a for loop with i to find first and last occurrence
    // Your code here
}

Hint

Loop over indices with for i in 0..temperatures.len(). Check if temperatures[i] == target. Use first.is_none() to set first once. Update last every time.

4

Print the results

Print the first and last occurrence indices using println! in this exact format: First occurrence: X and Last occurrence: Y. Use unwrap() on first and last.

DSA Rust

fn main() {
    let temperatures = vec![23, 25, 23, 22, 25, 23, 24];
    let target = 23;
    let mut first: Option = None;
    let mut last: Option = None;
    for i in 0..temperatures.len() {
        if temperatures[i] == target {
            if first.is_none() {
                first = Some(i);
            }
            last = Some(i);
        }
    }
    // Print first and last occurrence
    // Your code here
}

Hint

Use println!("First occurrence: {}", first.unwrap()); and similarly for last.

Practice

1. What does the "First and Last Occurrence of Element" problem ask you to find in a list?

easy

A. The first and last positions where a given element appears

B. The sum of all elements in the list

C. The maximum and minimum values in the list

D. The number of unique elements in the list

Solution

1. Step 1: Understand the problem statement

   The problem asks to find the first and last positions of a specific element in a list.

2. Step 2: Identify what is required

   We need to locate the earliest and latest index where the element appears.

3. Final Answer:

   The first and last positions where a given element appears -> Option A

4. Quick Check:

   Positions = first and last occurrence [OK]

Hint: Look for earliest and latest index of element [OK]

Common Mistakes:

• Confusing with sum or count of elements
• Looking for max/min values instead
• Ignoring the order of occurrences

2. Which Rust code snippet correctly finds the first occurrence index of element target in vector v?

easy

A. v.iter().position(|&x| x == target)

B. v.iter().find(|&&x| x == target)

C. v.iter().filter(|&&x| x == target).count()

D. v.iter().last().position(|&x| x == target)

Solution

1. Step 1: Understand Rust iterator methods

   The position method returns the index of the first element matching the condition.

2. Step 2: Check each option

   v.iter().position(|&x| x == target) uses position correctly to find the first occurrence index. v.iter().find(|&&x| x == target) returns an Option<&T>, not index. v.iter().filter(|&&x| x == target).count() counts matches, not index. v.iter().last().position(|&x| x == target) is invalid because last() returns an element, not an iterator.

3. Final Answer:

   v.iter().position(|&x| x == target) -> Option A

4. Quick Check:

   position() returns first index [OK]

Hint: Use .position() to get first index [OK]

Common Mistakes:

• Using find() which returns element, not index
• Using count() which returns number of matches
• Trying to call position() after last() which is invalid

3. What is the output of this Rust code?

let v = vec![2, 3, 5, 3, 7];
let target = 3;
let first = v.iter().position(|&x| x == target);
let last = v.iter().rposition(|&x| x == target);
println!("[{:?}, {:?}]", first, last);

medium

A. [Some(2), Some(3)]

B. [Some(1), Some(4)]

C. [None, None]

D. [Some(1), Some(3)]

Solution

1. Step 1: Find first occurrence of 3 in vector

   Vector is [2, 3, 5, 3, 7]. The first 3 is at index 1.

2. Step 2: Find last occurrence of 3 in vector

   The last 3 is at index 3.

3. Final Answer:

   [Some(1), Some(3)] -> Option D

4. Quick Check:

   First = 1, Last = 3 [OK]

Hint: Use position() and rposition() for first and last [OK]

Common Mistakes:

• Confusing indices of first and last occurrences
• Using wrong iterator methods
• Forgetting that indices start at 0

4. Identify the error in this Rust code that tries to find the last occurrence of target in vector v:

let last = v.iter().position(|&x| x == target).unwrap();
println!("{}", last);

medium

A. unwrap() causes panic if element not found

B. v.iter() cannot be used with position()

C. Using position() finds first, not last occurrence

D. Missing semicolon after unwrap()

Solution

1. Step 1: Understand what position() returns

   position() returns the index of the first element matching the condition, not the last.

2. Step 2: Identify correct method for last occurrence

   To find the last occurrence, rposition() should be used instead of position().

3. Final Answer:

   Using position() finds first, not last occurrence -> Option C

4. Quick Check:

   position() = first occurrence, not last [OK]

Hint: Use rposition() for last occurrence, not position() [OK]

Common Mistakes:

• Using position() for last occurrence
• Ignoring unwrap() panic possibility
• Thinking iter() can't use position()

5. Given a sorted vector v with possible duplicates, which approach efficiently finds the first and last occurrence of target using binary search in Rust?

hard

A. Use find() method twice with different closures

B. Use two binary searches: one for first occurrence, one for last occurrence

C. Use a single linear scan from start to end

D. Use position() and rposition() on the entire vector

Solution

1. Step 1: Understand the vector is sorted with duplicates

   Binary search can be used to find positions efficiently in sorted data.

2. Step 2: Use two binary searches for first and last occurrence

   One binary search finds the first index where target appears, another finds the last index.

3. Final Answer:

   Use two binary searches: one for first occurrence, one for last occurrence -> Option B

4. Quick Check:

   Binary search twice for first and last [OK]

Hint: Binary search twice for first and last occurrence [OK]

Common Mistakes:

• Using linear scan on large sorted vector
• Using position() which is linear time
• Trying to find both occurrences in one binary search