Binary Search Recursive Approach in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Binary Search Recursive Approach?

It divides the sorted list into two halves and recursively searches the half where the target value could be, reducing the search space by half each time.

beginner

In the recursive binary search, what are the base cases?

1. When the start index is greater than the end index (target not found).
2. When the middle element is equal to the target (target found).

beginner

Why must the input array be sorted for binary search to work?

Because binary search compares the target with the middle element to decide which half to search next, sorting ensures the halves are ordered and the search logic is valid.

beginner

What is the time complexity of the Binary Search Recursive Approach?

O(log n), where n is the number of elements in the array, because the search space halves with each recursive call.

intermediate

Show the recursive call pattern for searching target 7 in array [1,3,5,7,9].

1. Check middle element 5 (index 2): 7 > 5, search right half.
2. Check middle element 7 (index 3): found target.
Recursive calls reduce search space from 5 to 2 elements, then to 1 element.

What happens if the target is less than the middle element in recursive binary search?

A Return -1 immediately

B Search the right half of the array

C Search the left half of the array

D Search the entire array again

If the target is less than the middle element, binary search continues recursively on the left half.

What is the stopping condition when the target is not found in recursive binary search?

A When start index is greater than end index

B When array is empty

C When middle element equals target

D When start index equals end index

The search stops when the start index becomes greater than the end index, meaning the target is not in the array.

Which of these is required for binary search to work correctly?

A Array must be sorted

B Array must have unique elements

C Array must be unsorted

D Array must be reversed

Binary search requires the array to be sorted to decide which half to search next.

What is the time complexity of recursive binary search?

A O(1)

B O(n)

C O(n log n)

D O(log n)

Binary search reduces the search space by half each time, leading to O(log n) time complexity.

In recursive binary search, what parameters are typically passed in each recursive call?

A Array, target only

B Array, target, start index, end index

C Array, target, middle index

D Target, start index, end index

Each recursive call passes the array, target, and updated start and end indices to narrow the search.

Explain how the recursive binary search algorithm works step-by-step.

Think about how the array is split and how recursion narrows down the search.

Describe the base cases in recursive binary search and why they are important.

Base cases prevent infinite recursion and signal success or failure.

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