Binary Search Recursive Approach in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Binary Search on Sorted Array

What is the output of the following Java code that performs a recursive binary search for the value 7 in the array?

DSA Java

public class BinarySearch {
    public static int binarySearch(int[] arr, int left, int right, int target) {
        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, left, mid - 1, target);
        } else {
            return binarySearch(arr, mid + 1, right, target);
        }
    }

    public static void main(String[] args) {
        int[] array = {1, 3, 5, 7, 9, 11};
        int result = binarySearch(array, 0, array.length - 1, 7);
        System.out.println(result);
    }
}

A 4

B 2

C -1

D 3

💡 Hint

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

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

❓ Predict Output

intermediate

Result when target is not in the array

What will be printed when searching for the value 8 in the same array using the recursive binary search?

DSA Java

public class BinarySearch {
    public static int binarySearch(int[] arr, int left, int right, int target) {
        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, left, mid - 1, target);
        } else {
            return binarySearch(arr, mid + 1, right, target);
        }
    }

    public static void main(String[] args) {
        int[] array = {1, 3, 5, 7, 9, 11};
        int result = binarySearch(array, 0, array.length - 1, 8);
        System.out.println(result);
    }
}

A 3

B -1

C 4

D 0

💡 Hint

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

Since 8 is not in the array, the function returns -1 indicating the target is absent.

🧠 Conceptual

advanced

Understanding Recursive Binary Search Base Case

In the recursive binary search function, what is the purpose of the condition 'if (left > right)'?

A It checks if the target is found at the middle index.

B It resets the search to the full array again.

C It stops recursion when the search space is empty, meaning the target is not found.

D It moves the search to the right half of the array.

💡 Hint

Think about what happens when the search boundaries cross.

When left index is greater than right, it means the search space is empty and the target is not present.

🔧 Debug

advanced

Identify the error in this recursive binary search code

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

DSA Java

public class BinarySearch {
    public static int binarySearch(int[] arr, int left, int right, int target) {
        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, left, mid - 1, target);
        } else {
            return binarySearch(arr, mid + 1, right, target);
        }
    }

    public static void main(String[] args) {
        int[] array = {2, 4, 6, 8, 10};
        int result = binarySearch(array, 0, array.length - 1, 10);
        System.out.println(result);
    }
}

A Returns -1 even if target is present

B ArrayIndexOutOfBoundsException

C Infinite recursion leading to StackOverflowError

D Correct output: 4

💡 Hint

Check the base case condition carefully.

The base case uses 'left >= right' which excludes the case when left == right, missing the last element check, so it returns -1 even if target is present.

🚀 Application

expert

Number of Recursive Calls for Binary Search

Given a sorted array of 64 elements, what is the maximum number of recursive calls made by a recursive binary search to find a target or conclude it is not present?

A 7

B 8

C 6

D 9

💡 Hint

Think about how many times you can divide 64 by 2 until you reach 1.

Binary search divides the search space by half each call. For 64 elements, maximum calls = log2(64) + 1 = 6 + 1 = 7 calls. But since the first call counts as call 1, the maximum depth is 7 calls. However, the question asks for maximum number of recursive calls, which is 7 calls including the first. So answer is 7.

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