Binary Search as Divide and Conquer in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Binary Search Call

What is the output of the following Rust code that performs a recursive binary search for the value 7 in a sorted array?

DSA Rust

fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> {
    if low > high {
        return None;
    }
    let mid = low + (high - low) / 2;
    if arr[mid] == target {
        Some(mid)
    } else if arr[mid] > target {
        binary_search(arr, target, low, mid - 1)
    } else {
        binary_search(arr, target, mid + 1, high)
    }
}

fn main() {
    let arr = [1, 3, 5, 7, 9, 11];
    let result = binary_search(&arr, 7, 0, arr.len() - 1);
    println!("{:?}", result);
}

A None

B Some(2)

C Some(3)

D Some(4)

💡 Hint

Remember that array indices start at 0 and the array is sorted.

The target 7 is at index 3 in the array [1, 3, 5, 7, 9, 11]. The binary search finds it and returns Some(3).

🧠 Conceptual

intermediate

Understanding Divide and Conquer in Binary Search

Which statement best describes how binary search uses the divide and conquer approach?

A It searches both halves of the array simultaneously to find the target faster.

B It searches the array linearly but divides the target value into smaller parts.

C It divides the array into multiple parts and searches all parts in parallel.

D It divides the array into two halves and recursively searches only the half that may contain the target.

💡 Hint

Think about how the search space changes after each comparison.

Binary search splits the array into two halves and only continues searching the half where the target could be, reducing the search space each time.

🔧 Debug

advanced

Identify the Bug in Binary Search Implementation

What error will occur when running this Rust binary search code and why?

DSA Rust

fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> {
    if low > high {
        return None;
    }
    let mid = low + (high - low) / 2;
    if arr[mid] == target {
        Some(mid)
    } else if arr[mid] > target {
        binary_search(arr, target, low, mid - 1)
    } else {
        binary_search(arr, target, mid + 1, high)
    }
}

fn main() {
    let arr = [2, 4, 6, 8, 10];
    let result = binary_search(&arr, 1, 0, arr.len() - 1);
    println!("{:?}", result);
}

A Panic due to unsigned integer underflow when mid is 0 and mid - 1 is computed

B Returns None even if the target is present

C Infinite recursion causing stack overflow

D Compiles but returns wrong index for target

💡 Hint

Check the base case condition and what happens when mid is zero.

When mid is 0, mid - 1 causes unsigned integer underflow, leading to panic at runtime.

🚀 Application

advanced

Binary Search on a Rotated Sorted Array

Given a rotated sorted array, which approach correctly finds the target using binary search?

A Always search the left half first, then the right half if not found

B Check which half is sorted, then decide which half to search next based on target position

C Perform linear search because binary search does not work on rotated arrays

D Sort the array first, then perform binary search

💡 Hint

In a rotated array, one half is always sorted.

By checking which half is sorted, we can decide where the target might be and continue binary search accordingly.

❓ Predict Output

expert

Output of Iterative Binary Search with Edge Cases

What is the output of this Rust code performing iterative binary search for target 15 in the array?

DSA Rust

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

fn main() {
    let arr = [5, 10, 15, 20, 25];
    let result = binary_search_iter(&arr, 15);
    println!("{:?}", result);
}

A Some(2)

B Some(3)

C None

D Some(1)

💡 Hint

Remember that array indices start at 0 and the target is 15.

The target 15 is at index 2. The iterative binary search returns Some(2).

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