Binary Search Recursive Approach in DSA C# - Build from Scratch

Binary Search Recursive Approach

📖 Scenario: You are working on a library system that needs to quickly find if a book ID exists in a sorted list of book IDs.To do this efficiently, you will use the binary search algorithm implemented with recursion.

🎯 Goal: Build a recursive binary search function in C# that checks if a given book ID is present in a sorted array of book IDs.

📋 What You'll Learn

Create a sorted integer array called bookIds with the exact values: 101, 203, 305, 407, 509, 611, 713

Create two integer variables start and end to represent the search range

Write a recursive function called BinarySearch that takes bookIds, start, end, and target as parameters and returns a boolean indicating if target is found

Call BinarySearch with the full range and the target book ID 407

Print "Found" if the book ID is found, otherwise print "Not Found"

💡 Why This Matters

🌍 Real World

Binary search is used in many applications like searching in databases, libraries, and any sorted data to quickly find information.

💼 Career

Understanding binary search and recursion is important for software developers to write efficient search algorithms and optimize performance.

1

Create the sorted array of book IDs

Create a sorted integer array called bookIds with these exact values: 101, 203, 305, 407, 509, 611, 713

DSA C#

// Create the sorted array of book IDs
// Your code here

Hint

Use int[] bookIds = { ... }; to create the array with the exact values.

2

Set up the start and end indices for the search

Create two integer variables called start and end. Set start to 0 and end to the last index of bookIds.

DSA C#

int[] bookIds = {101, 203, 305, 407, 509, 611, 713};
// Create start and end variables
// Your code here

Hint

Set start to 0 and end to bookIds.Length - 1.

3

Write the recursive binary search function

Write a recursive function called BinarySearch that takes parameters int[] bookIds, int start, int end, and int target. The function should return true if target is found in bookIds between start and end, otherwise false. Use the binary search logic with recursion.

DSA C#

int[] bookIds = {101, 203, 305, 407, 509, 611, 713};
int start = 0;
int end = bookIds.Length - 1;

// Write the recursive BinarySearch function below
// Your code here

Hint

Use the middle index to compare with target. Call BinarySearch recursively on the left or right half.

4

Call the function and print the result

Call the BinarySearch function with bookIds, start, end, and the target value 407. Print "Found" if the function returns true, otherwise print "Not Found".

DSA C#

int[] bookIds = {101, 203, 305, 407, 509, 611, 713};
int start = 0;
int end = bookIds.Length - 1;

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

// Call BinarySearch with target 407 and print the result
// Your code here

Hint

Call BinarySearch with the target 407. Use an if statement to print "Found" or "Not Found".

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