Binary Search as Divide and Conquer in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the middle index in binary search.

DSA Rust

let mid = (low + [1]) / 2;

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

Amid

Blow

C0

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high.

Using mid variable before defining it.

The middle index is calculated by adding low and high, then dividing by 2.

2fill in blank

medium

Complete the code to check if the middle element is equal to the target.

DSA Rust

if arr[mid] == [1] {

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Amid

Btarget

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with mid index instead of target value.

Using low or high instead of target.

We compare the middle element with the target value we are searching for.

3fill in blank

hard

Fix the error in updating the low index when target is greater than middle element.

DSA Rust

if arr[mid] < target {
    low = [1];
}

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Amid + 1

Bhigh

Clow + 1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid instead of mid + 1.

Using high or low + 1 incorrectly.

When target is greater, we move low to mid + 1 to search the right half.

4fill in blank

hard

Fill both blanks to update the high index when target is less than middle element.

DSA Rust

if arr[mid] > target {
    high = [1];
} else {
    low = [2];
}

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

Amid - 1

Bmid + 1

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to mid instead of mid - 1.

Setting low to mid instead of mid + 1.

If target is less, high moves to mid - 1; else low moves to mid + 1.

5fill in blank

hard

Fill all three blanks to complete the binary search function signature and return statement.

DSA Rust

fn [1](arr: &[i32], target: i32) -> [2] {
    let mut low = 0;
    let mut high = arr.len() as i32 - 1;
    while low <= high {
        let mid = (low + high) / 2;
        if arr[mid as usize] == target {
            return [3];
        }
        // rest of code omitted
    }
    -1
}

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

Abinary_search

Bi32

Cmid as i32

Dusize

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong function name.

Returning wrong type or variable.

The function is named binary_search, returns i32 index, and returns mid as i32 when found.

Practice

1. What is the main idea behind Binary Search as a divide and conquer algorithm?

easy

A. Sort the list first, then search linearly

B. Check every element one by one until the target is found

C. Repeatedly split the sorted list in half to find the target

D. Use a hash table to find the target instantly

Solution

1. Step 1: Understand divide and conquer in binary search

   Binary search works by dividing the sorted list into two halves and deciding which half to search next.

2. Step 2: Identify the correct approach

   It repeatedly splits the list in half, narrowing down the search space until the target is found or no elements remain.

3. Final Answer:

   Repeatedly split the sorted list in half to find the target -> Option C

4. Quick Check:

   Divide and conquer = split list halves [OK]

Hint: Binary search halves the list each step [OK]

Common Mistakes:

• Thinking binary search checks every element
• Assuming list must be unsorted
• Confusing binary search with linear search

2. Which of the following is the correct Rust function signature for a binary search function that returns an Option<usize>?

easy

A. fn binary_search(arr: &[i32], target: i32) -> Option<usize>

B. fn binary_search(arr: Vec bool

C. fn binary_search(arr: &[i32], target: i32) -> usize

D. fn binary_search(arr: &mut [i32], target: i32) -> i32

Solution

1. Step 1: Understand Rust function signature for binary search

   The function takes a slice of integers and a target integer, returning an Option with the index if found.

2. Step 2: Match the correct signature

   Option<usize> is used to return Some(index) if found or None if not found, matching fn binary_search(arr: &[i32], target: i32) -> Option<usize>.

3. Final Answer:

   fn binary_search(arr: &[i32], target: i32) -> Option<usize> -> Option A

4. Quick Check:

   Return Option<usize> for index or None [OK]

Hint: Use Option<usize> to handle found or not found [OK]

Common Mistakes:

• Returning usize without Option causes errors if not found
• Using Vec instead of slice for input unnecessarily
• Returning bool loses index information

3. What will be the output of this Rust code snippet using binary search?

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

fn main() {
    let nums = [2, 4, 6, 8, 10];
    println!("{:?}", binary_search(&nums, 6));
}

medium

A. Some(3)

B. Some(2)

C. None

D. Some(1)

Solution

1. Step 1: Trace binary search on array [2,4,6,8,10] for target 6

   Initial left=0, right=4, mid=2. arr[2]=6 matches target.

2. Step 2: Return index 2 wrapped in Some

   Since arr[mid] == target, function returns Some(2).

3. Final Answer:

   Some(2) -> Option B

4. Quick Check:

   Index of 6 in array is 2 [OK]

Hint: Check mid index value equals target first [OK]

Common Mistakes:

• Off-by-one errors in mid calculation
• Confusing index with value
• Returning None incorrectly when target exists

4. Identify the bug in this binary search Rust code and choose the correct fix:

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

medium

A. Change while condition to left < right

B. Change mid calculation to (left + right) / 2 without subtraction

C. Remove else if and use only if and else

D. Change right initialization to arr.len() - 1

Solution

1. Step 1: Analyze right boundary initialization

   Right is set to arr.len(), which is out of bounds since indexing is 0-based.

2. Step 2: Fix right to arr.len() - 1 to avoid out-of-bounds access

   Setting right to arr.len() - 1 ensures mid index stays within valid range.

3. Final Answer:

   Change right initialization to arr.len() - 1 -> Option D

4. Quick Check:

   Right boundary must be last valid index [OK]

Hint: Right index = length - 1 for zero-based arrays [OK]

Common Mistakes:

• Using arr.len() as right causes panic
• Changing while condition incorrectly
• Incorrect mid calculation causing overflow

5. Given a sorted array with possible duplicate values, how can you modify binary search to find the first occurrence of a target value?

hard

A. After finding target at mid, continue searching left half to find earlier occurrence

B. Return immediately when target is found at mid

C. Search right half only after finding target

D. Use linear search after binary search to find first occurrence

Solution

1. Step 1: Understand duplicates and first occurrence requirement

   Binary search normally returns any matching index, but to find first occurrence, we must check if earlier duplicates exist.

2. Step 2: Modify search to continue left after finding target

   When target is found at mid, record index and continue searching left half to find if an earlier occurrence exists.

3. Final Answer:

   After finding target at mid, continue searching left half to find earlier occurrence -> Option A

4. Quick Check:

   Find first occurrence by searching left after match [OK]

Hint: Keep searching left half after finding target [OK]

Common Mistakes:

• Stopping search immediately on first match
• Searching right half instead of left
• Using linear search unnecessarily