Count Occurrences of Element in Sorted Array in DSA Rust - Time & Space Complexity

Time Complexity: Count Occurrences of Element in Sorted Array

O(log n)

Understanding Time Complexity

We want to know how fast we can count how many times a number appears in a sorted list.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn count_occurrences(arr: &[i32], target: i32) -> usize {
    fn find_first(arr: &[i32], target: i32) -> Option {
        let mut left = 0;
        let mut right = arr.len();
        while left < right {
            let mid = (left + right) / 2;
            if arr[mid] < target {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        if left < arr.len() && arr[left] == target { Some(left) } else { None }
    }

    fn find_last(arr: &[i32], target: i32) -> Option {
        let mut left = 0;
        let mut right = arr.len();
        while left < right {
            let mid = (left + right) / 2;
            if arr[mid] <= target {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        if right > 0 && arr[right - 1] == target { Some(right - 1) } else { None }
    }

    match (find_first(arr, target), find_last(arr, target)) {
        (Some(first), Some(last)) => last - first + 1,
        _ => 0,
    }
}
    

This code finds the first and last place of the target number using binary search, then counts how many times it appears.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two binary search loops to find first and last positions.
• How many times: Each binary search loop runs about log₂(n) times, where n is the array size.

How Execution Grows With Input

As the list gets bigger, the number of steps grows slowly because each search cuts the list in half.

Input Size (n)	Approx. Operations
10	About 2 x 4 = 8 steps
100	About 2 x 7 = 14 steps
1000	About 2 x 10 = 20 steps

Pattern observation: The steps grow slowly, roughly doubling the input size adds only one more step per search.

Final Time Complexity

Time Complexity: O(log n)

This means the time to count occurrences grows slowly as the list gets bigger, making it very efficient.

Common Mistake

[X] Wrong: "Counting occurrences requires checking every element one by one, so it must be O(n)."

[OK] Correct: Because the list is sorted, we can use binary search to jump directly to the target's start and end, avoiding checking all elements.

Interview Connect

Knowing how to use binary search to count occurrences shows you understand how to use sorting to speed up searches, a key skill in many coding challenges.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

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