Binary Search Recursive Approach in DSA Javascript - Build from Scratch

Binary Search Recursive Approach

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

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

📋 What You'll Learn

Create a sorted array called bookIDs with the exact values: [101, 203, 305, 407, 509, 611, 713]

Create a variable called targetID and set it to 407

Write a recursive function called binarySearch that takes arr, target, start, and end as parameters

Use recursion to find if targetID is in bookIDs

Print true if found, otherwise false

💡 Why This Matters

🌍 Real World

Binary search is used in many applications like searching in databases, looking up words in dictionaries, or finding items in sorted lists quickly.

💼 Career

Understanding binary search helps in software development roles that require efficient data searching and algorithm optimization.

1

Create the sorted array of book IDs

Create a sorted array called bookIDs with these exact values: [101, 203, 305, 407, 509, 611, 713]

DSA Javascript

// Create the sorted array bookIDs with the given values
// Your code here

Hint

Use const bookIDs = [101, 203, 305, 407, 509, 611, 713]; to create the array.

2

Set the target book ID to search

Create a variable called targetID and set it to 407

DSA Javascript

const bookIDs = [101, 203, 305, 407, 509, 611, 713];
// Create the variable targetID and set it to 407
// Your code here

Hint

Use const targetID = 407; to store the book ID you want to find.

3

Write the recursive binary search function

Write a recursive function called binarySearch that takes parameters arr, target, start, and end. It should return true if target is found in arr, otherwise false. Use recursion to check the middle element and search the left or right half accordingly.

DSA Javascript

const bookIDs = [101, 203, 305, 407, 509, 611, 713];
const targetID = 407;

// Write the recursive binarySearch function below
// Your code here

Hint

Use the middle index to compare and call binarySearch recursively on the left or right half.

4

Print the result of searching for the target ID

Use console.log to print the result of calling binarySearch with bookIDs, targetID, 0 as start, and bookIDs.length - 1 as end.

DSA Javascript

const bookIDs = [101, 203, 305, 407, 509, 611, 713];
const targetID = 407;

function binarySearch(arr, target, start, end) {
    if (start > end) {
        return false;
    }
    const mid = Math.floor((start + end) / 2);
    if (arr[mid] === target) {
        return true;
    } else if (arr[mid] > target) {
        return binarySearch(arr, target, start, mid - 1);
    } else {
        return binarySearch(arr, target, mid + 1, end);
    }
}

// Print the result of binarySearch with bookIDs, targetID, 0, and bookIDs.length - 1
// Your code here

Hint

Call binarySearch(bookIDs, targetID, 0, bookIDs.length - 1) and print the result.

Practice

1. What is the main idea behind the binary search recursive approach when searching for a value in a sorted array?

easy

A. Divide the array into halves and search recursively in the half where the value can be.

B. Check every element one by one from start to end.

C. Sort the array first and then search linearly.

D. Use a loop to check elements 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: Recognize recursive approach

   In recursion, the function calls itself 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 A

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Remember: binary search splits array, recursion repeats on half [OK]

Common Mistakes:

• Thinking binary search checks every element linearly
• Confusing recursion with iteration
• Not requiring the array to be sorted

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

easy

A. if (array.length == 0) return 0;

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

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

D. if (mid == target) return mid + 1;

Solution

1. Step 1: Identify base case meaning

   The base case stops recursion when the search space is empty, which happens if start index passes end index.

2. Step 2: Match base case condition

   The condition start > end means no elements left to search, so return -1 to indicate not found.

3. Final Answer:

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

4. Quick Check:

   Base case = empty search space [OK]

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

Common Mistakes:

• Using incorrect comparison like start == end as base case
• Returning wrong index or value in base case
• Confusing base case with recursive step

3. What will be the output of the following JavaScript code?

function binarySearch(arr, target, start, end) {
  if (start > end) return -1;
  const mid = Math.floor((start + end) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, target, start, mid - 1);
  else return binarySearch(arr, target, mid + 1, end);
}

const array = [2, 4, 6, 8, 10];
console.log(binarySearch(array, 6, 0, array.length - 1));

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Trace initial call

   Start=0, end=4, mid=2 (element 6). Since arr[mid] == target (6), return mid = 2.

2. Step 2: Confirm output

   The function returns 2 immediately without further recursion.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

Hint: Check mid index value first to find target quickly [OK]

Common Mistakes:

• Miscomputing mid index
• Confusing index with value
• Forgetting zero-based indexing

4. Identify the error in the following recursive binary search function and choose the correct fix.

function binarySearch(arr, target, start, end) {
  if (start >= end) return -1;
  const mid = Math.floor((start + end) / 2);
  if (arr[mid] === target) return mid;
  else if (arr[mid] > target) return binarySearch(arr, target, start, mid - 1);
  else return binarySearch(arr, target, mid + 1, end);
}

medium

A. Change mid calculation to (start + end + 1) / 2;

B. Remove else if and use only if statements.

C. Return mid + 1 when found instead of mid;

D. Change base case to if (start > end) return -1;

Solution

1. Step 1: Analyze base case condition

   The base case incorrectly uses start >= end, which misses the case when start == end and the element could still be at that index.

2. Step 2: Correct base case

   Changing it to start > end ensures the search includes the last element before stopping.

3. Final Answer:

   Change base case to if (start > end) return -1; -> Option D

4. Quick Check:

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

Hint: Base case should allow checking when start equals end [OK]

Common Mistakes:

• Stopping search too early by wrong base case
• Changing mid calculation unnecessarily
• Returning wrong index on success

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. Search right half only when arr[mid] == target.

B. Return mid immediately when arr[mid] == target without further search.

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

D. Use linear search after binary search to find first occurrence.

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any occurrence, but to find first occurrence, we must check if earlier duplicates exist.

2. Step 2: Modify recursion on match

   When arr[mid] == target, recursively search left half to find if an earlier occurrence exists; if none, mid is first occurrence.

3. Final Answer:

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

4. Quick Check:

   Find first occurrence by searching left on match [OK]

Hint: On match, search left half to find first duplicate [OK]

Common Mistakes:

• Returning first found index without checking left side
• Searching right half on match incorrectly
• Using linear search unnecessarily