Binary Search Recursive Approach in DSA Java - Build from Scratch

Binary Search Recursive Approach

📖 Scenario: You have a sorted list of book IDs in a library system. You want to quickly find if a specific book ID is available using a fast search method.

🎯 Goal: Build a recursive binary search function in Java that checks if a given book ID exists in the sorted list.

📋 What You'll Learn

Create an integer array called bookIDs with the exact values {101, 203, 305, 407, 509, 611, 713}

Create an integer variable called targetID and set it to 407

Write a recursive method called binarySearch that takes parameters int[] arr, int left, int right, and int target

The binarySearch method should return true if target is found in arr, otherwise false

Call binarySearch with bookIDs, 0, bookIDs.length - 1, and targetID

Print the result of the search

💡 Why This Matters

🌍 Real World

Binary search is used in many software systems to quickly find data in sorted lists, such as searching for user IDs, product codes, or book IDs in libraries.

💼 Career

Understanding binary search and recursion is fundamental for software developers, especially in roles involving algorithms, data processing, and performance optimization.

1

Create the sorted array of book IDs

Create an integer array called bookIDs with the exact values {101, 203, 305, 407, 509, 611, 713}

DSA Java

# Your code here

Hint

Use int[] bookIDs = {101, 203, 305, 407, 509, 611, 713}; inside the main method.

2

Add the target book ID to search

Create an integer variable called targetID and set it to 407 inside the main method, below the bookIDs array

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] bookIDs = {101, 203, 305, 407, 509, 611, 713};
        // Your code here
    }
}

Hint

Declare int targetID = 407; inside the main method.

3

Write the recursive binary search method

Write a recursive method called binarySearch that takes parameters int[] arr, int left, int right, and int target. It should return true if target is found in arr, otherwise false. Place this method inside the Main class but outside the main method.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] bookIDs = {101, 203, 305, 407, 509, 611, 713};
        int targetID = 407;
        // Your code here
    }

    // Write binarySearch method here
}

Hint

Use the base case if (left > right) return false;. Calculate mid and compare arr[mid] with target. Call binarySearch recursively on the left or right half.

4

Call binarySearch and print the result

Inside the main method, call binarySearch with bookIDs, 0, bookIDs.length - 1, and targetID. Store the result in a boolean variable called found. Then print found.

DSA Java

public class Main {
    public static void main(String[] args) {
        int[] bookIDs = {101, 203, 305, 407, 509, 611, 713};
        int targetID = 407;
        // Your code here
    }

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

Hint

Call binarySearch(bookIDs, 0, bookIDs.length - 1, targetID), save the result in found, then print 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. 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