Why Count Occurrences of Element in Sorted Array in DSA Rust?

The Big Idea

What if you could count items in a huge sorted list instantly without checking each one?

The Scenario

Imagine you have a long list of sorted numbers, like a phone book sorted by last names. You want to find out how many times a particular name appears. Doing this by checking each name one by one would take a lot of time.

The Problem

Manually scanning through the entire list to count occurrences is slow and tiring. It's easy to make mistakes, especially if the list is very long. You might miss some entries or count some twice.

The Solution

Using a smart method that takes advantage of the sorted order lets you quickly find where the element starts and ends. This way, you count occurrences fast without checking every item.

Before vs After

✗ Before

let mut count = 0;
for &item in &sorted_array {
    if item == target {
        count += 1;
    }
}
println!("Count: {}", count);

✓ After

let first = find_first_occurrence(&sorted_array, target);
let last = find_last_occurrence(&sorted_array, target);
let count = if first == -1 { 0 } else { last - first + 1 };
println!("Count: {}", count);

What It Enables

This lets you quickly find how many times an element appears in a sorted list, even if the list is huge.

Real Life Example

Counting how many times a word appears in a sorted dictionary or how many customers bought a specific product from a sorted sales list.

Key Takeaways

Manual counting is slow and error-prone.

Using the sorted order helps find occurrences quickly.

Efficient counting saves time and reduces mistakes.

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