Binary Search Iterative Approach in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index in the binary search loop.

DSA Rust

let mut left = 0;
let mut right = arr.len() as i32 - 1;
while left <= right {
    let mid = [1];
    // rest of the code
}

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

A(left + right) / 2

Bleft + right

Cleft - right

Dright - left

Attempts:

3 left

💡 Hint

Common Mistakes

Using left + right without dividing by 2 causes index out of range.

Subtracting indices instead of averaging.

The middle index is calculated as the average of left and right indices: (left + right) / 2.

2fill in blank

medium

Complete the code to compare the middle element with the target value.

DSA Rust

if arr[mid as usize] [1] target {
    return Some(mid as usize);
} else if arr[mid as usize] < target {
    left = mid + 1;
} else {
    right = mid - 1;
}

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

A>

B!=

C==

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > or < instead of == causes wrong matches.

Using != leads to incorrect logic.

We check if the middle element equals the target to find the element.

3fill in blank

hard

Fix the error in updating the left pointer when the middle element is less than the target.

DSA Rust

if arr[mid as usize] < target {
    [1] = mid + 1;
}

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

Aleft

Bright

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Updating right instead of left causes infinite loop.

Assigning mid or target instead of left.

When middle element is less than target, move left pointer to mid + 1 to search right half.

4fill in blank

hard

Fill both blanks to correctly update the right pointer when the middle element is greater than the target.

DSA Rust

if arr[mid as usize] > target {
    [1] = [2] - 1;
}

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

Aright

Bleft

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Updating left instead of right causes wrong search direction.

Using target instead of mid causes logic errors.

When middle element is greater than target, move right pointer to mid - 1 to search left half.

5fill in blank

hard

Fill all three blanks to complete the iterative binary search function returning the index or None.

DSA Rust

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

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

A(left + right) / 2

Bmid as usize

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Returning wrong index type causes errors.

Updating wrong pointer causes infinite loops.

Calculate mid as average, return mid as usize when found, and update left to mid + 1 when target is greater.

Practice

1. What is the main purpose of the binary search iterative approach in a sorted array?

easy

A. To find the position of a target value efficiently by repeatedly dividing the search interval in half

B. To sort the array elements in ascending order

C. To reverse the array elements

D. To find the maximum value in the array

Solution

1. Step 1: Understand binary search goal

   Binary search aims to find a target value's position in a sorted array quickly by dividing the search range repeatedly.

2. Step 2: Compare with other options

   Sorting, reversing, or finding max are different operations unrelated to binary search's main goal.

3. Final Answer:

   To find the position of a target value efficiently by repeatedly dividing the search interval in half -> Option A

4. Quick Check:

   Binary search = divide and conquer search [OK]

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

Common Mistakes:

• Confusing binary search with sorting algorithms
• Thinking binary search works on unsorted arrays
• Assuming binary search finds max or min values

2. Which of the following is the correct Rust syntax to start the iterative binary search loop over a sorted slice arr with indices low and high?

easy

A. while low <= high { /* loop body */ }

B. for i in low..high { /* loop body */ }

C. loop { if low > high { break; } /* loop body */ }

D. while low < high { /* loop body */ }

Solution

1. Step 1: Identify correct loop condition for binary search

   Binary search continues while low index is less than or equal to high index to cover all elements.

2. Step 2: Compare options

   while low <= high { /* loop body */ } uses while low <= high, which is correct. for i in low..high { /* loop body */ } uses a for loop which is not suitable. loop { if low > high { break; } /* loop body */ } uses an infinite loop with break, which is less clear. while low < high { /* loop body */ } uses low < high, which misses the case when low equals high.

3. Final Answer:

   while low <= high { /* loop body */ } -> Option A

4. Quick Check:

   Loop condition includes equality for last element [OK]

Hint: Use while low <= high to cover all elements [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using < instead of <= causing missed element check
• Using infinite loop without proper break condition

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

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

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

medium

A. -1

B. 1

C. 2

D. 3

Solution

1. Step 1: Trace binary search for target 6 in array

   Array is [2,4,6,8,10]. Initial low=0, high=4. mid=2. arr[2]=6 equals target, so return 2.

2. Step 2: Confirm returned index

   Index 2 is correct position of 6 in the array.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Target 6 found at index 2 [OK]

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

Common Mistakes:

• Miscomputing mid index causing wrong return
• Returning -1 incorrectly when target exists
• Confusing index with value in output

4. Identify the error in this Rust iterative binary search function and choose the fix:

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

medium

A. Change while condition to low < high

B. Remove type casting of mid to usize

C. Change mid calculation to mid = low - (high - low) / 2

D. Change high initialization to arr.len() as i32 - 1

Solution

1. Step 1: Analyze high initialization

   High is set to arr.len() which is out of bounds index (since last valid index is length - 1).

2. Step 2: Correct high initialization

   Set high to arr.len() - 1 to avoid index out of bounds error during access.

3. Final Answer:

   Change high initialization to arr.len() as i32 - 1 -> Option D

4. Quick Check:

   High must be last valid index [OK]

Hint: High index = length - 1 to avoid out of bounds [OK]

Common Mistakes:

• Setting high to length causing panic
• Changing while condition incorrectly
• Incorrect mid calculation formula

5. Given a sorted array arr = [1, 3, 5, 7, 9, 11], write an iterative binary search function in Rust to find the index of the smallest element greater than or equal to a target value. What will be the output of your function when searching for target 8?

hard

A. 5

B. 4

C. -1

D. 3

Solution

1. Step 1: Understand problem - find smallest element >= target

   We want the index of the first element not less than target 8.

2. Step 2: Trace binary search for target 8

   Array: [1,3,5,7,9,11]. Elements >=8 are 9 (index 4) and 11 (index 5). The smallest is 9 at index 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   First element >= 8 is 9 at index 4 [OK]

Hint: Find first element >= target by adjusting high on match [OK]

Common Mistakes:

• Returning index of element just less than target
• Returning -1 when element exists
• Confusing smallest element greater than target with equal