Binary Search Recursive Approach in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

int mid = low [1] (high - low) / 2;

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

A-

B*

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition

Using multiplication or division incorrectly

The middle index is found by adding low to half the distance between high and low.

2fill in blank

medium

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

DSA C#

if (arr[mid] [1] target) return mid;

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A==

B!=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment operator '=' instead of '=='

Using inequality operators instead of equality

We check if the middle element is exactly equal to the target to find the element.

3fill in blank

hard

Fix the error in the recursive call when the target is less than the middle element.

DSA C#

return BinarySearch(arr, low, [1], target);

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

Amid + 1

Bmid - 1

Chigh - 1

Dhigh + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 which searches the right half incorrectly

Changing low instead of high

If the target is less than the middle element, search the left half by setting high to mid - 1.

4fill in blank

hard

Fill both blanks to complete the recursive call when the target is greater than the middle element.

DSA C#

return BinarySearch(arr, [1], [2], target);

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

Amid + 1

Blow

Chigh

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping low and high values

Using mid - 1 instead of mid + 1

When the target is greater than the middle element, search the right half by setting low to mid + 1 and keeping high unchanged.

5fill in blank

hard

Fill all three blanks to complete the base case that stops recursion when the element is not found.

DSA C#

if ([1] > [2]) return [3];

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

Alow

Bhigh

C-1

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Returning mid instead of -1

Checking high > low instead of low > high

The search stops when low is greater than high, meaning the target is not in the array, so return -1.

Practice

1. What is the main idea behind the Binary Search Recursive Approach when searching for a value in a sorted array?

easy

A. Use a queue to store elements and search iteratively.

B. Check every element one by one from start to end.

C. Sort the array first and then search linearly.

D. Divide the array into halves and search recursively in the half where the value can be.

Solution

1. Step 1: Understand binary search logic

   Binary search works by dividing the sorted array into two halves and deciding which half to search next based on the middle element.

2. Step 2: Recognize recursive approach

   The recursive approach calls the same search function on the half where the target value may exist, reducing the search space each time.

3. Final Answer:

   Divide the array into halves and search recursively in the half where the value can be. -> Option D

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Remember: binary search splits array, searches half recursively [OK]

Common Mistakes:

• Thinking binary search checks every element linearly
• Not realizing array must be sorted first
• Confusing iterative and recursive approaches

2. Which of the following is the correct base case condition in a recursive binary search function in C#?

easy

A. if (array[start] == target) return start;

B. if (start == end) return start;

C. if (start > end) return -1;

D. if (end < start) return 0;

Solution

1. Step 1: Identify base case meaning

   The base case stops recursion when the search space is invalid, meaning start index is greater than end index.

2. Step 2: Match condition to code

   The condition if (start > end) return -1; correctly indicates the target is not found.

3. Final Answer:

   if (start > end) return -1; -> Option C

4. Quick Check:

   Base case = invalid range returns -1 [OK]

Hint: Base case: start index passes end index means not found [OK]

Common Mistakes:

• Using wrong comparison like start == end as base case
• Returning 0 which can be a valid index
• Not handling base case causing infinite recursion

3. What will be the output of the following C# code snippet?

int[] arr = {2, 4, 6, 8, 10};
int result = BinarySearch(arr, 0, arr.Length - 1, 6);
Console.WriteLine(result);

int BinarySearch(int[] array, int start, int end, int target) {
    if (start > end) return -1;
    int mid = (start + end) / 2;
    if (array[mid] == target) return mid;
    else if (array[mid] > target) return BinarySearch(array, start, mid - 1, target);
    else return BinarySearch(array, mid + 1, end, target);
}

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Trace initial call and mid calculation

   Initial call: start=0, end=4, mid=(0+4)/2=2. array[2]=6 matches target 6.

2. Step 2: Return mid index

   Since array[mid] == target, function returns 2 immediately.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Check mid index first; if match, return immediately [OK]

Common Mistakes:

• Miscomputing mid index
• Returning wrong index due to off-by-one errors
• Confusing array values with indices

4. Identify the error in this recursive binary search function and choose the correct fix:

int BinarySearch(int[] array, int start, int end, int target) {
    if (start >= end) return -1;
    int mid = (start + end) / 2;
    if (array[mid] == target) return mid;
    else if (array[mid] > target) return BinarySearch(array, start, mid - 1, target);
    else return BinarySearch(array, mid + 1, end, target);
}

medium

A. Change mid calculation to mid = (start + end + 1) / 2;

B. Change base case to if (start > end) return -1;

C. Return 0 instead of -1 in base case.

D. Remove the base case entirely.

Solution

1. Step 1: Analyze base case condition

   The base case uses start >= end, which wrongly excludes the case when start == end, missing the last element check.

2. Step 2: Correct base case to allow checking when start == end

   Changing to start > end ensures the last element is checked before stopping recursion.

3. Final Answer:

   Change base case to if (start > end) return -1; -> Option B

4. Quick Check:

   Base case must allow start == end check [OK]

Hint: Base case should be start > end, not start >= end [OK]

Common Mistakes:

• Stopping recursion too early with wrong base case
• Changing mid calculation unnecessarily
• Returning 0 which can be a valid index

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

hard

A. When array[mid] == target, continue searching left half to find earlier occurrence.

B. Return mid immediately when array[mid] == target.

C. Search right half only when array[mid] == target.

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

Solution

1. Step 1: Understand duplicates and first occurrence

   Duplicates mean target can appear multiple times; to find first occurrence, we must check if earlier indices also have the target.

2. Step 2: Modify recursion to search left half when target found

   When array[mid] == target, recursively search left half to find if target appears earlier; if none found, return current mid.

3. Final Answer:

   When array[mid] == target, continue searching left half to find earlier occurrence. -> Option A

4. Quick Check:

   Find first occurrence by searching left half on match [OK]

Hint: On match, search left half to find first occurrence [OK]

Common Mistakes:

• Returning first found index without checking left side
• Searching right half on match which finds last occurrence
• Using linear search after binary search unnecessarily