Why Binary Search and What Sorted Order Gives You in DSA Javascript - See It Work

Why Binary Search and What Sorted Order Gives You

📖 Scenario: Imagine you have a sorted list of book titles in a library catalog. You want to find if a certain book is available quickly without checking every title one by one.

🎯 Goal: You will build a simple program that uses binary search on a sorted list of book titles to find if a specific book is in the catalog.

📋 What You'll Learn

Create a sorted array of book titles

Set a target book title to search for

Implement binary search logic to find the target

Print whether the book was found or not

💡 Why This Matters

🌍 Real World

Binary search is used in many apps and systems to quickly find data in sorted lists, like searching contacts, files, or products.

💼 Career

Understanding binary search and sorted data is essential for software developers to write efficient search functions and optimize performance.

1

Create a sorted array of book titles

Create a variable called books and assign it the sorted array ["Algorithms", "Data Structures", "JavaScript Basics", "Python Essentials", "Web Development"].

DSA Javascript

// Your code here

Hint

Remember to keep the array sorted as given.

2

Set the target book title to search for

Create a variable called target and assign it the string "Python Essentials".

DSA Javascript

const books = ["Algorithms", "Data Structures", "JavaScript Basics", "Python Essentials", "Web Development"];
// Your code here

Hint

Use the exact string "Python Essentials" for the target.

3

Implement binary search logic

Write a function called binarySearch that takes arr and target as parameters. Use variables left, right, and mid to perform binary search on the sorted array. Return true if target is found, otherwise false.

DSA Javascript

const books = ["Algorithms", "Data Structures", "JavaScript Basics", "Python Essentials", "Web Development"];
const target = "Python Essentials";

// Your code here

Hint

Use a while loop to narrow down the search range until you find the target or the range is empty.

4

Print whether the book was found

Use console.log to print "Found" if binarySearch(books, target) returns true, otherwise print "Not Found".

DSA Javascript

const books = ["Algorithms", "Data Structures", "JavaScript Basics", "Python Essentials", "Web Development"];
const target = "Python Essentials";

function binarySearch(arr, target) {
    let left = 0;
    let right = arr.length - 1;
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);
        if (arr[mid] === target) {
            return true;
        } else if (arr[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    return false;
}

// Your code here

Hint

Use an if-else statement to print the correct message based on the search result.

Practice

1. Why is binary search only applicable on a sorted list?

easy

A. Because it can search any list regardless of order

B. Because it relies on comparing the middle element to decide which half to search next

C. Because it checks every element one by one

D. Because it sorts the list before searching

Solution

1. Step 1: Understand binary search logic

   Binary search works by comparing the middle element to the target and deciding which half to continue searching.

2. Step 2: Importance of sorted order

   This logic only works if the list is sorted, so we know which half contains the target based on comparison.

3. Final Answer:

   Because it relies on comparing the middle element to decide which half to search next -> Option B

4. Quick Check:

   Binary search needs sorted list = Because it relies on comparing the middle element to decide which half to search next [OK]

Hint: Binary search splits sorted list by middle comparison [OK]

Common Mistakes:

• Thinking binary search works on unsorted lists
• Confusing binary search with linear search
• Assuming binary search sorts the list automatically

2. Which of the following is the correct syntax to find the middle index in a binary search on array arr between indices low and high in JavaScript?

easy

A. const mid = Math.floor((low + high) / 2);

B. const mid = (low + high) / 2;

C. const mid = Math.ceil((low + high) / 2);

D. const mid = low + high;

Solution

1. Step 1: Calculate middle index correctly

   The middle index should be the floor of the average of low and high to get an integer index.

2. Step 2: Check each option

   const mid = Math.floor((low + high) / 2); uses Math.floor which correctly rounds down. const mid = (low + high) / 2; returns a float, const mid = Math.ceil((low + high) / 2); rounds up which can cause errors, const mid = low + high; sums indices without division.

3. Final Answer:

   const mid = Math.floor((low + high) / 2); -> Option A

4. Quick Check:

   Use Math.floor for middle index [OK]

Hint: Use Math.floor to get integer middle index [OK]

Common Mistakes:

• Not using Math.floor causing float index
• Using Math.ceil causing off-by-one errors
• Adding indices without dividing

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

const arr = [2, 4, 6, 8, 10];
function binarySearch(array, target) {
  let low = 0, high = array.length - 1;
  while (low <= high) {
    const mid = Math.floor((low + high) / 2);
    if (array[mid] === target) return mid;
    else if (array[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}
console.log(binarySearch(arr, 8));

medium

A. 3

B. 2

C. -1

D. 4

Solution

1. Step 1: Trace binary search steps for target 8

   Initial low=0, high=4, mid=2 (value 6). Since 6 < 8, low = mid+1 = 3.

2. Step 2: Next iteration mid=3 (value 8), matches target, return 3

   The function returns index 3 where 8 is found.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Index of 8 in array = 3 [OK]

Hint: Trace mid updates to find target index [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in mid calculation
• Returning -1 incorrectly

4. The following binary search code has a bug. What is the error?

function binarySearch(arr, target) {
  let low = 0;
  let high = arr.length;
  while (low <= high) {
    let mid = Math.floor((low + high) / 2);
    if (arr[mid] === target) return mid;
    else if (arr[mid] < target) low = mid + 1;
    else high = mid - 1;
  }
  return -1;
}

medium

A. The loop condition should be low < high, not low <= high

B. low should start at 1, not 0

C. mid calculation should use Math.ceil instead of Math.floor

D. high should be initialized to arr.length - 1, not arr.length

Solution

1. Step 1: Check initialization of high

   Array indices go from 0 to length-1, so high must be arr.length - 1 to avoid out-of-bounds.

2. Step 2: Consequences of wrong high value

   Setting high = arr.length causes mid to sometimes be arr.length, which is invalid index and causes errors.

3. Final Answer:

   high should be initialized to arr.length - 1, not arr.length -> Option D

4. Quick Check:

   High index = length - 1 [OK]

Hint: High index must be last valid array index [OK]

Common Mistakes:

• Setting high to arr.length causing index errors
• Changing low start incorrectly
• Changing loop condition wrongly

5. You have a sorted array arr = [1, 3, 3, 5, 7, 7, 7, 9]. You want to find the first occurrence of the number 7 using binary search. Which modification helps achieve this?

hard

A. Use linear search instead of binary search

B. Return mid immediately when arr[mid] equals target

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

D. Search right half when arr[mid] equals target

Solution

1. Step 1: Understand duplicate elements in sorted array

   Binary search normally returns any matching index, but to find first occurrence, we must check if earlier indices also have the target.

2. Step 2: Modify search to continue left when target found

   When arr[mid] equals target, update result but continue searching left half to find earlier occurrence.

3. Final Answer:

   When arr[mid] equals target, continue searching left half to find first occurrence -> Option C

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 immediately on first match
• Searching right half on match
• Using linear search unnecessarily