Binary Search Recursive Approach in DSA C# - Time & Space Complexity

Time Complexity: Binary Search Recursive Approach

O(log n)

Understanding Time Complexity

We want to understand how the time taken by recursive binary search changes as the input size grows.

Specifically, how many steps does it take to find an item in a sorted list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int BinarySearch(int[] arr, int target, int left, int right)
{
    if (left > right) return -1;
    int mid = left + (right - left) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] > target) return BinarySearch(arr, target, left, mid - 1);
    else return BinarySearch(arr, target, mid + 1, right);
}
    

This code searches for a target value in a sorted array by repeatedly dividing the search range in half.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call that halves the search range each time.
• How many times: At most, the function calls itself until the range is empty, roughly halving the size each time.

How Execution Grows With Input

Each recursive step cuts the search area in half, so the number of steps grows slowly as the input size grows.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: Doubling the input size adds only one extra step.

Final Time Complexity

Time Complexity: O(log n)

This means the number of steps grows slowly, increasing by one step each time the input size doubles.

Common Mistake

[X] Wrong: "Binary search checks every element one by one, so it is O(n)."

[OK] Correct: Binary search skips half the elements each step, so it does not check all elements, making it much faster than linear search.

Interview Connect

Understanding binary search's time complexity shows you can analyze efficient algorithms that use divide and conquer, a key skill in many coding challenges.

Self-Check

"What if the array was not sorted? How would the time complexity of searching change?"

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