First and Last Occurrence of Element in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the first occurrence index of target in the array.

DSA Rust

fn first_occurrence(arr: &[i32], target: i32) -> Option<usize> {
    for i in 0..arr.len() {
        if arr[i] == [1] {
            return Some(i);
        }
    }
    None
}

Drag options to blanks, or click blank then click option'

Aarr[i]

Bi

C0

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using the index variable instead of the target value in the comparison.

Comparing arr[i] to 0 instead of the target.

We compare each element arr[i] with the target value to find the first occurrence.

2fill in blank

medium

Complete the code to find the last occurrence index of target in the array.

DSA Rust

fn last_occurrence(arr: &[i32], target: i32) -> Option<usize> {
    let mut result = None;
    for i in 0..arr.len() {
        if arr[i] == [1] {
            result = Some(i);
        }
    }
    result
}

Drag options to blanks, or click blank then click option'

Atarget

Bi

Carr[i]

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[i] to the index i instead of the target.

Assigning result to None inside the loop.

We update result whenever we find an element equal to the target, so the last occurrence is stored.

3fill in blank

hard

Fix the error in the code to correctly find the first occurrence of target using a while loop.

DSA Rust

fn first_occurrence_while(arr: &[i32], target: i32) -> Option<usize> {
    let mut i = 0;
    while i < arr.len() {
        if arr[i] == [1] {
            return Some(i);
        }
        i += 1;
    }
    None
}

Drag options to blanks, or click blank then click option'

Aarr[i]

Bi

Ctarget

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[i] to the index i instead of the target.

Using a wrong variable in the if condition.

The comparison must be between the current element arr[i] and the target value.

4fill in blank

hard

Fill both blanks to create a function that returns a tuple with first and last occurrence indices of target.

DSA Rust

fn first_and_last(arr: &[i32], target: i32) -> (Option<usize>, Option<usize>) {
    let mut first = None;
    let mut last = None;
    for i in 0..arr.len() {
        if arr[i] == [1] {
            if first.is_none() {
                first = Some(i);
            }
            last = [2];
        }
    }
    (first, last)
}

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Atarget

BSome(i)

Ci

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using i instead of Some(i) for last.

Comparing arr[i] to i instead of target.

We check if arr[i] equals target, set first if not set, and update last to Some(i) each time.

5fill in blank

hard

Fill all three blanks to create a function that returns a tuple with first and last occurrence indices using iterators and enumerate.

DSA Rust

fn first_and_last_iter(arr: &[i32], target: i32) -> (Option<usize>, Option<usize>) {
    let mut first = None;
    let mut last = None;
    for (i, &val) in arr.iter().enumerate() {
        if val == [1] {
            if first == [2] {
                first = Some(i);
            }
            last = [3];
        }
    }
    (first, last)
}

Drag options to blanks, or click blank then click option'

Atarget

BNone

CSome(i)

DSome(val)

Attempts:

3 left

💡 Hint

Common Mistakes

Using Some(i) instead of None in the if condition.

Assigning last to Some(val) instead of Some(i).

We compare val to target, check if first is None, and update last to Some(i).

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