Binary Search Recursive Approach in DSA Java - 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 a number 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; // not found
    }
    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 number 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: The search range is cut in half until the target is found or range is empty.

How Execution Grows With Input

Each step cuts the problem size roughly in half, so the number of steps grows slowly as 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 does not check all elements; it splits the search range in half each time, so it only needs a few steps even for large inputs.

Interview Connect

Understanding binary search time complexity shows you can analyze efficient searching methods, a key skill in many coding challenges and real-world problems.

Self-Check

"What if we changed the recursive binary search to an iterative version? How would the time complexity 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. Check every element one by one from start to end

B. Sort the array first and then search linearly

C. Divide the array into halves and search recursively in the half where the value may exist

D. Use a stack to store all elements and pop until the value is found

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 comparison.

2. Step 2: Apply recursion in binary search

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

3. Final Answer:

   Divide the array into halves and search recursively in the half where the value may exist -> Option C

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Binary search splits array, search half recursively [OK]

Common Mistakes:

• Thinking binary search checks every element
• Not using recursion or iteration to reduce search space
• Assuming array is unsorted

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

easy

A. if (start < end) return -1;

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

C. if (array[mid] == target) return mid;

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

Solution

1. Step 1: Identify base case for recursion end

   The base case ends recursion when the search space is invalid, i.e., start index is greater than end index.

2. Step 2: Confirm correct base case syntax

   Using if (start > end) return -1; correctly signals the target is not found.

3. Final Answer:

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

4. Quick Check:

   Base case = start > end [OK]

Hint: Base case ends when start passes end [OK]

Common Mistakes:

• Using start == end as base case incorrectly
• Returning wrong index when target not found
• Confusing base case with target found condition

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

public class Test {
  public static int binarySearch(int[] arr, int start, int end, int target) {
    if (start > end) return -1;
    int mid = start + (end - start) / 2;
    if (arr[mid] == target) return mid;
    else if (arr[mid] > target) return binarySearch(arr, start, mid - 1, target);
    else return binarySearch(arr, mid + 1, end, target);
  }
  public static void main(String[] args) {
    int[] data = {2, 4, 6, 8, 10};
    System.out.println(binarySearch(data, 0, data.length - 1, 6));
  }
}

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Trace initial call and mid calculation

   Array is [2,4,6,8,10], start=0, end=4, target=6. Mid = 0 + (4-0)/2 = 2.

2. Step 2: Check if arr[mid] equals target

   arr[2] = 6 equals target 6, so return mid = 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Calculate mid, compare, return index if match [OK]

Common Mistakes:

• Miscomputing mid index
• Returning wrong index when target found
• Confusing array indices

4. Identify the error in the following recursive binary search code snippet:

public static int binarySearch(int[] arr, int start, int end, int target) {
  if (start >= end) return -1;
  int mid = (start + end) / 2;
  if (arr[mid] == target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, start, mid - 1, target);
  else return binarySearch(arr, mid + 1, end, target);
}

medium

A. Base case condition should be start > end, not start >= end

B. Mid calculation should be start - end instead of start + end

C. Should return mid + 1 when target is greater

D. No error, code is correct

Solution

1. Step 1: Analyze base case condition

   Base case currently is if (start >= end) return -1;. This causes missing the case when start == end and target is at that index.

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

   Changing to if (start > end) return -1; ensures the single element at start==end is checked.

3. Final Answer:

   Base case condition should be start > end, not start >= end -> Option A

4. Quick Check:

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

Hint: Base case excludes start == end incorrectly [OK]

Common Mistakes:

• Using >= instead of > in base case
• Ignoring single element check
• Incorrect mid calculation

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. Return mid immediately when arr[mid] == target

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

C. Search only the right half when arr[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 target.

2. Step 2: Modify recursive call when target found

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

3. Final Answer:

   When arr[mid] == target, continue searching left half to find earlier occurrence -> Option B

4. Quick Check:

   Search left half on match for first occurrence [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
• Using linear search unnecessarily