Binary Search Iterative Approach in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

int mid = (low [1] high) / 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 instead of addition.

Dividing low by 2 before adding high.

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

2fill in blank

medium

Complete the condition to check if the middle element is the target.

DSA C#

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

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

A!=

B<

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than or greater than instead of equality.

Using not equal operator which is incorrect here.

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

3fill in blank

hard

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

DSA C#

if (arr[mid] < target) low = [1] + 1;

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

Atarget

Blow

Cmid

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to low + 1 instead of mid + 1.

Setting low to high or target which is incorrect.

When middle element is less than target, we move low to mid + 1 to search the right half.

4fill in blank

hard

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

DSA C#

if (arr[mid] > target) [1] = [2] - 1;

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

Ahigh

Blow

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using low instead of high.

Subtracting one from low or target instead of mid.

When middle element is greater than target, we move high to mid - 1 to search the left half.

5fill in blank

hard

Fill all three blanks to complete the binary search loop condition and return statement.

DSA C#

while ([1] <= [2]) {
    int mid = (low + high) / 2;
    if (arr[mid] == target) return [3];
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
}
return -1;

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

Alow

Bhigh

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using target in loop condition instead of high.

Returning target instead of mid index.

Using wrong variables in loop condition.

The loop runs while low is less than or equal to high. If found, return mid index.

Practice

1. What is the main purpose of the Binary Search Iterative Approach in a sorted array?

easy

A. To find the maximum value in the array

B. To sort the array elements in ascending order

C. To reverse the array elements

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

Solution

1. Step 1: Understand binary search goal

   Binary search aims to find a target value's position quickly by checking the middle element and narrowing the search range.

2. Step 2: Identify iterative approach

   The iterative method uses a loop to repeatedly divide the search interval until the target is found or the interval is empty.

3. Final Answer:

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

4. Quick Check:

   Binary search = efficient target search [OK]

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

Common Mistakes:

• Confusing binary search with sorting
• Thinking it reverses the array
• Assuming it finds max value

2. Which of the following is the correct way to calculate the middle index in an iterative binary search in C# to avoid overflow?

easy

A. int mid = low + (high - low) / 2;

B. int mid = (low + high) / 2;

C. int mid = (low - high) / 2;

D. int mid = (low * high) / 2;

Solution

1. Step 1: Understand overflow risk

   Adding low and high directly can cause integer overflow if values are large.

2. Step 2: Use safe middle calculation

   Calculating mid as low + (high - low) / 2 avoids overflow by subtracting before adding.

3. Final Answer:

   int mid = low + (high - low) / 2; -> Option A

4. Quick Check:

   Safe mid calculation = low + (high - low) / 2 [OK]

Hint: Use low + (high - low) / 2 to prevent overflow [OK]

Common Mistakes:

• Using (low + high) / 2 causing overflow
• Subtracting incorrectly in mid calculation
• Multiplying indices instead of adding

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

int[] arr = {2, 4, 6, 8, 10};
int target = 6;
int low = 0, high = arr.Length - 1;
int result = -1;
while (low <= high) {
  int mid = low + (high - low) / 2;
  if (arr[mid] == target) {
    result = mid;
    break;
  } else if (arr[mid] < target) {
    low = mid + 1;
  } else {
    high = mid - 1;
  }
}
Console.WriteLine(result);

medium

A. -1

B. 3

C. 2

D. 0

Solution

1. Step 1: Trace the binary search steps

   Initial low=0, high=4, mid=2 (element 6). arr[2] == target 6, so result=2 and loop breaks.

2. Step 2: Output the result

   Since target found at index 2, Console.WriteLine prints 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Check middle element first to find target quickly [OK]

Common Mistakes:

• Confusing index with value
• Not updating low or high correctly
• Assuming target not found returns 0

4. Identify the error in the following iterative binary search code snippet and choose the correct fix:

int BinarySearch(int[] arr, int target) {
  int low = 0, high = arr.Length - 1;
  while (low < high) {
    int mid = low + (high - low) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}

medium

A. Change while condition to 'while (low <= high)'

B. Change mid calculation to '(low + high) / 2'

C. Remove 'return -1;' at the end

D. Change 'low = mid + 1;' to 'low = mid - 1;'

Solution

1. Step 1: Analyze loop condition

   The loop uses 'while (low < high)', which misses checking when low equals high, possibly skipping the last element.

2. Step 2: Correct loop condition

   Changing to 'while (low <= high)' ensures all elements are checked including when low equals high.

3. Final Answer:

   Change while condition to 'while (low <= high)' -> Option A

4. Quick Check:

   Loop must include low == high to check all elements [OK]

Hint: Use <= in loop to include last element check [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Incorrect mid calculation causing overflow
• Wrong update of low or high pointers

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

hard

A. Return immediately when target is found

B. When target is found, continue searching left half to find earlier occurrence

C. Search only the right half after finding target

D. Use linear search instead of binary search

Solution

1. Step 1: Understand duplicates in sorted array

   Duplicates mean target can appear multiple times; we want the first index where it appears.

2. Step 2: Modify binary search to find first occurrence

   When target found at mid, record mid but continue searching left half (high = mid - 1) to find earlier occurrence.

3. Final Answer:

   When target is found, continue searching left half to find earlier occurrence -> Option B

4. Quick Check:

   Find first occurrence by searching left after match [OK]

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

Common Mistakes:

• Stopping search immediately on first match
• Searching right half after finding target
• Switching to linear search unnecessarily