Binary Search Iterative Approach in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Search for Existing Element

What is the output of the following C# code that performs an iterative binary search for the value 7 in a sorted array?

DSA C#

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

int[] data = {1, 3, 5, 7, 9, 11};
int result = BinarySearch(data, 7);
System.Console.WriteLine(result);

A 3

B 2

C -1

D 4

💡 Hint

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

The value 7 is at index 3 in the array. The binary search finds it and returns the index.

❓ Predict Output

intermediate

Output of Binary Search for Non-Existing Element

What is the output of the following C# code that performs an iterative binary search for the value 8 in a sorted array?

DSA C#

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

int[] data = {2, 4, 6, 10, 12};
int result = BinarySearch(data, 8);
System.Console.WriteLine(result);

A -1

B 2

C 3

D 0

💡 Hint

If the target is not found, the function returns -1.

The value 8 is not in the array, so the function returns -1.

🧠 Conceptual

advanced

Why Use mid = left + (right - left) / 2 Instead of (left + right) / 2?

In the iterative binary search, why is the middle index calculated as mid = left + (right - left) / 2 instead of mid = (left + right) / 2?

A Because it is faster to compute

B To ensure mid is always rounded up

C To avoid integer overflow when left and right are large values

D To make the code compatible with floating point arrays

💡 Hint

Think about what happens if left and right are very large numbers.

Using (left + right) can cause integer overflow if left and right are large. The formula left + (right - left) / 2 avoids this by subtracting first.

🔧 Debug

advanced

Identify the Bug in Binary Search Implementation

What error will occur when running this iterative binary search code?

DSA C#

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

A NullReferenceException

B No error, works correctly

C Infinite loop

D IndexOutOfRangeException at arr[mid]

💡 Hint

Check the initial value of right and how it relates to array indices.

The right boundary is set to arr.Length, which is out of valid index range (0 to Length-1). Accessing arr[mid] can cause IndexOutOfRangeException.

🚀 Application

expert

Number of Iterations in Worst Case Binary Search

Given a sorted array of 1024 elements, what is the maximum number of iterations the iterative binary search algorithm will perform to find a target or conclude it is not present?

A 9

B 10

C 11

D 12

💡 Hint

Binary search halves the search space each iteration. Think about powers of 2.

The maximum iterations equal the smallest integer k where 2^k >= 1024. Since 2^10 = 1024, max iterations are 10.

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