Count Occurrences of Element in Sorted Array in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of counting occurrences in a sorted array

What is the output of the following Rust code that counts occurrences of the number 3 in a sorted array?

DSA Rust

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let first = arr.binary_search(&target).ok();
    if first.is_none() {
        return 0;
    }
    let first_index = first.unwrap();
    let mut count = 1;
    let mut i = first_index + 1;
    while i < arr.len() && arr[i] == target {
        count += 1;
        i += 1;
    }
    count
}

fn main() {
    let arr = [1, 2, 3, 3, 3, 4, 5];
    let result = count_occurrences(&arr, 3);
    println!("{}", result);
}

A 2

B 3

C 1

D 0

💡 Hint

Think about how many times 3 appears consecutively in the array.

The binary_search finds an occurrence of 3 at index 2. Then the loop counts how many times 3 appears after that index. Since 3 appears at indices 2, 3, and 4, the output is 3.

❓ Predict Output

intermediate

Output when element is not present

What will be printed when counting occurrences of 6 in the sorted array?

DSA Rust

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let first = arr.binary_search(&target).ok();
    if first.is_none() {
        return 0;
    }
    let first_index = first.unwrap();
    let mut count = 1;
    let mut i = first_index + 1;
    while i < arr.len() && arr[i] == target {
        count += 1;
        i += 1;
    }
    count
}

fn main() {
    let arr = [1, 2, 3, 3, 3, 4, 5];
    let result = count_occurrences(&arr, 6);
    println!("{}", result);
}

A panic

B 1

C 6

D 0

💡 Hint

If the element is not found, what does the function return?

binary_search returns None for elements not found, so the function returns 0 indicating no occurrences.

🧠 Conceptual

advanced

Why use binary search for counting occurrences?

Why is binary search a good choice for counting occurrences of an element in a sorted array?

A Because it finds the element in O(log n) time, allowing efficient counting afterwards.

B Because it sorts the array before counting.

C Because it counts all elements without searching.

D Because it uses recursion to count elements.

💡 Hint

Think about the speed of searching in sorted arrays.

Binary search quickly finds the target element's position in O(log n) time, which is faster than scanning the entire array.

❓ Predict Output

advanced

Output of counting occurrences with duplicates at start and end

What is the output of counting occurrences of 2 in this sorted array?

DSA Rust

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let first = arr.binary_search(&target).ok();
    if first.is_none() {
        return 0;
    }
    let first_index = first.unwrap();
    let mut count = 1;
    let mut i = first_index + 1;
    while i < arr.len() && arr[i] == target {
        count += 1;
        i += 1;
    }
    count
}

fn main() {
    let arr = [2, 2, 2, 3, 4, 5, 6, 2];
    let result = count_occurrences(&arr, 2);
    println!("{}", result);
}

A 4

B 1

C 3

D 0

💡 Hint

Remember the array is sorted except the last element.

binary_search finds a 2 at index 2. The loop counts how many times 2 appears after that index. Since arr[3] == 3 != 2, the output is 1. The last 2 at index 7 is not counted because it is not consecutive.

🔧 Debug

expert

Identify the bug in counting occurrences code

What is the bug in this Rust code for counting occurrences of an element in a sorted array?

DSA Rust

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let mut count = 0;
    for &item in arr {
        if item == target {
            count += 1;
        } else if count > 0 {
            break;
        }
    }
    count
}

fn main() {
    let arr = [1, 2, 2, 2, 3, 4, 5];
    let result = count_occurrences(&arr, 2);
    println!("{}", result);
}

A It assumes the array is sorted but does not use binary search, so it is less efficient.

B It counts all occurrences correctly but breaks early, so it misses some elements.

C It will panic if the target is not found.

D It counts occurrences only if they are at the start of the array.

💡 Hint

Consider the time complexity and the use of binary search.

The code scans the array linearly and breaks after counting consecutive occurrences, which works but is less efficient than binary search for large sorted arrays.

Practice

1. What is the main advantage of using binary search to count occurrences of an element in a sorted array?

easy

A. It counts elements without searching

B. It sorts the array automatically

C. It reduces the search time to logarithmic complexity

D. It works only on unsorted arrays

Solution

1. Step 1: Understand binary search efficiency

   Binary search divides the search space in half each time, making it very fast on sorted arrays.

2. Step 2: Compare with linear search

   Linear search checks each element, which is slower (linear time). Binary search is faster (logarithmic time).

3. Final Answer:

   It reduces the search time to logarithmic complexity -> Option C

4. Quick Check:

   Binary search = O(log n) time [OK]

Hint: Binary search halves search space each step [OK]

Common Mistakes:

• Thinking binary search sorts the array
• Assuming binary search works on unsorted arrays
• Confusing counting with searching

2. Which Rust code snippet correctly declares a function to count occurrences of target in a sorted slice arr?

easy

A. fn count_occurrences(arr: &[i32], target: i32) -> usize { /* code */ }

B. fn count_occurrences(arr: Vec<i32>, target: i32) -> i32 { /* code */ }

C. fn count_occurrences(arr: &[i32], target: i32) -> i32 { /* code */ }

D. fn count_occurrences(arr: Vec<i32>, target: i32) -> usize { /* code */ }

Solution

1. Step 1: Check parameter types

   Using a slice &[i32] is efficient for read-only access without ownership transfer.

2. Step 2: Check return type

   Counting occurrences returns a non-negative number, so usize is appropriate.

3. Final Answer:

   fn count_occurrences(arr: &[i32], target: i32) -> usize { /* code */ } -> Option A

4. Quick Check:

   Slice + usize return = correct signature [OK]

Hint: Use slice &[i32] and usize for count [OK]

Common Mistakes:

• Using Vec when slice is enough
• Returning i32 instead of usize for counts
• Confusing ownership with borrowing

3. What is the output of this Rust code counting occurrences of 3 in the sorted array?

fn main() {
    let arr = [1, 2, 3, 3, 3, 3, 5];
    let count = count_occurrences(&arr, 3);
    println!("{}", count);
}

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let first = arr.binary_search(&target).unwrap_or_else(|x| x);
    let mut count = 0;
    for &num in &arr[first..] {
        if num == target {
            count += 1;
        } else {
            break;
        }
    }
    count
}

medium

A. 4

B. 3

C. 1

D. 0

Solution

1. Step 1: Find occurrence index

   binary_search returns Ok(index) for 3 at index 2.

2. Step 2: Count consecutive 3s from index 2

   Elements at indices 2,3,4,5 are 3, so count = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   4 occurrences counted correctly [OK]

Hint: Start counting from binary_search index [OK]

Common Mistakes:

• Using unwrap_or_else returns insertion point, not found index
• Counting beyond target elements
• Not breaking loop on different element

4. Identify the bug in this Rust function that counts occurrences of target in a sorted array arr:

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    let start = arr.binary_search(&target).unwrap_or_else(|_| 0);
    let mut count = 0;
    for &num in &arr[start..] {
        if num == target {
            count += 1;
        } else {
            break;
        }
    }
    count
}

medium

A. Function returns i32 instead of usize

B. The loop does not break on different element

C. binary_search returns index of last occurrence, not first

D. unwrap_or_else(|_| 0) returns 0 even if target not found, causing wrong count

Solution

1. Step 1: Understand unwrap_or_else behavior

   unwrap_or_else(|_| 0) returns 0 if target not found, which is incorrect start index.

2. Step 2: Consequence of wrong start index

   Counting from 0 when target is absent counts wrong elements, giving wrong count.

3. Final Answer:

   unwrap_or_else(|_| 0) returns 0 even if target not found, causing wrong count -> Option D

4. Quick Check:

   unwrap_or_else should use insertion point or handle not found [OK]

Hint: Use unwrap_or_else with insertion point for not found [OK]

Common Mistakes:

• Using unwrap_or_else(|| 0)
• Not breaking loop on different element
• Confusing first and last occurrence index

5. Given a sorted array arr with possible duplicates, which Rust approach best counts occurrences of target efficiently and correctly for all cases, including when target is absent? Options: A) Use binary_search to find insertion points of target and target+1, then subtract indices. B) Use binary_search to find any occurrence, then count forward until different element. C) Use linear scan from start to end counting matches. D) Use binary_search to find last occurrence, then count backward until different element.

hard

A. Find one occurrence with binary_search, count forward until different element

B. Find insertion points for target and target+1, subtract to get count

C. Linear scan entire array counting matches

D. Find last occurrence with binary_search, count backward until different element

Solution

1. Step 1: Understand insertion points method

   Finding insertion points for target and target+1 gives start and end indices of target range.

2. Step 2: Calculate count by subtraction

   Subtracting these indices gives exact count, even if target is absent (count = 0).

3. Step 3: Compare with other methods

   Counting forward or backward after binary_search can fail if target not found; linear scan is slow.

4. Final Answer:

   Find insertion points for target and target+1, subtract to get count -> Option B

5. Quick Check:

   Insertion points subtraction = accurate count [OK]

Hint: Use binary_search for insertion points, subtract indices [OK]

Common Mistakes:

• Counting forward without checking if target found
• Using linear scan on large arrays
• Counting backward without binary_search for last occurrence