Why Binary Search as Divide and Conquer in DSA Typescript?

The Big Idea

What if you could find a needle in a haystack by looking at only a few spots?

The Scenario

Imagine you have a huge phone book and you want to find a friend's number. You start from the first page and check every name one by one until you find it.

The Problem

This manual way is very slow and tiring because you might have to look through thousands of pages. It wastes time and energy, especially if the book is very big.

The Solution

Binary Search splits the phone book in half each time you look. You check the middle page, decide which half your friend's name is in, and then only search that half. This way, you quickly zoom in on the right page without checking every single one.

Before vs After

✗ Before

function findNumber(names: string[], target: string): number {
  for (let i = 0; i < names.length; i++) {
    if (names[i] === target) return i;
  }
  return -1;
}

✓ After

function binarySearch(names: string[], target: string): number {
  let left = 0;
  let right = names.length - 1;
  while (left <= right) {
    const mid = Math.floor((left + right) / 2);
    if (names[mid] === target) return mid;
    else if (names[mid] < target) left = mid + 1;
    else right = mid - 1;
  }
  return -1;
}

What It Enables

It lets you find things super fast in big sorted lists by cutting the search space in half again and again.

Real Life Example

When you search for a word in a dictionary app, it uses binary search to jump quickly to the right page instead of flipping through every word.

Key Takeaways

Manual search checks every item one by one, which is slow.

Binary Search divides the list to find the target faster.

This method works only if the list is sorted.

Practice

1. What is the main idea behind Binary Search as a divide and conquer algorithm?

easy

A. Check every element one by one until the target is found

B. Use a hash table to find the target instantly

C. Sort the list first and then search linearly

D. Repeatedly split the sorted list into halves to find the target

Solution

1. Step 1: Understand binary search principle

   Binary search works by dividing the sorted list into two halves and deciding which half to continue searching in.

2. Step 2: Compare target with middle element

   By comparing the target with the middle element, we eliminate half the list each time, making search efficient.

3. Final Answer:

   Repeatedly split the sorted list into halves to find the target -> Option D

4. Quick Check:

   Divide and conquer = split list halves [OK]

Hint: Divide sorted list in halves to find target fast [OK]

Common Mistakes:

• Thinking binary search works on unsorted lists
• Searching linearly instead of dividing
• Ignoring the middle element comparison

2. Which of the following is the correct TypeScript function signature for a binary search function?

easy

A. function binarySearch(arr: number[], target: number): number

B. function binarySearch(arr: number, target: number[]): number

C. function binarySearch(arr: string[], target: string): void

D. function binarySearch(arr: number[], target: number): void

Solution

1. Step 1: Identify input and output types

   Binary search takes an array of numbers and a target number, returning the index (number) or -1 if not found.

2. Step 2: Match function signature

   function binarySearch(arr: number[], target: number): number matches input types (number array and number) and returns a number index, which is correct.

3. Final Answer:

   function binarySearch(arr: number[], target: number): number -> Option A

4. Quick Check:

   Input: number[], number; Output: number [OK]

Hint: Input array and target number, output index number [OK]

Common Mistakes:

• Swapping array and target types
• Returning void instead of index
• Using wrong array element types

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

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

const nums = [2, 4, 6, 8, 10];
console.log(binarySearch(nums, 6));

medium

A. 2

B. 3

C. -1

D. 1

Solution

1. Step 1: Identify array and target

   Array is [2,4,6,8,10], target is 6.

2. Step 2: Trace binary search steps

   Middle index initially is 2 (value 6), which matches target, so function returns 2 immediately.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Index of 6 in array = 2 [OK]

Hint: Find middle index and compare target [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in mid calculation
• Returning wrong index when target found

4. Identify the bug in this binary search implementation and choose the fix:

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

medium

A. Change mid calculation to (left + right + 1) / 2

B. Change while condition to left < right

C. Change right initialization to arr.length - 1

D. Remove else if and use only if statements

Solution

1. Step 1: Check right boundary initialization

   Right should be last valid index, which is arr.length - 1, not arr.length (out of bounds).

2. Step 2: Understand impact on loop

   Using arr.length causes mid to sometimes be out of array range, causing runtime errors.

3. Final Answer:

   Change right initialization to arr.length - 1 -> Option C

4. Quick Check:

   Right boundary = last index [OK]

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

Common Mistakes:

• Setting right to arr.length causes out-of-bounds
• Changing loop condition incorrectly
• Miscalculating mid index

5. You have a sorted array with duplicate values: [1, 2, 2, 2, 3, 4]. How would you modify binary search to find the first occurrence of 2?

hard

A. Search the right half after finding 2

B. After finding 2, continue searching left half to find earlier occurrence

C. Return the first found index immediately

D. Use linear search instead of binary search

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any matching index, not necessarily the first occurrence.

2. Step 2: Modify search to find first occurrence

   After finding target, continue searching left half to check if earlier occurrence exists, updating result accordingly.

3. Final Answer:

   After finding 2, continue searching left half to find earlier occurrence -> Option B

4. Quick Check:

   Find first occurrence by searching left side [OK]

Hint: Keep searching left after match to find first duplicate [OK]

Common Mistakes:

• Stopping at first found index
• Searching right half after match
• Switching to linear search unnecessarily