Binary Search as Divide and Conquer in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the divide and conquer approach in binary search?

Divide the problem into smaller parts by splitting the array into halves, then solve the smaller part where the target might be, repeating until found or empty.

beginner

In binary search, what condition tells you to search the left half of the array?

If the target value is less than the middle element, search the left half because the array is sorted.

beginner

Why does binary search require the array to be sorted?

Because it decides which half to search based on comparisons, sorting ensures the target is only in one half, making the search efficient.

intermediate

What is the time complexity of binary search and why?

O(log n) because each step cuts the search space in half, quickly reducing the number of elements to check.

intermediate

Show the recursive step in binary search as divide and conquer.

Check middle element; if target equals middle, return index; else recursively search left or right half depending on comparison.

What does binary search do at each step?

A Splits the array into two halves and searches one half

B Checks every element one by one

C Sorts the array first

D Searches both halves simultaneously

Binary search divides the array into halves and searches only the half where the target could be.

Which condition means you should search the right half in binary search?

A Target is less than middle element

B Target is equal to middle element

C Target is greater than middle element

D Array is unsorted

If the target is greater than the middle element, binary search continues in the right half.

What is the base case for the recursive binary search?

A When the target is greater than all elements

B When the array is sorted

C When the array size is odd

D When the array is empty or target is found

The recursion stops when the target is found or the search space is empty.

Why is binary search faster than linear search on sorted arrays?

A Because it checks elements randomly

B Because it halves the search space each step

C Because it sorts the array first

D Because it uses more memory

Binary search halves the search space each step, reducing the number of checks quickly.

What must be true about the array for binary search to work correctly?

A It must be sorted

B It must have unique elements

C It must be unsorted

D It must be empty

Binary search requires a sorted array to decide which half to search.

Explain how binary search uses divide and conquer to find a target in a sorted array.

Think about cutting the problem size in half each time.

Describe the base case and recursive step in a recursive binary search function.

Focus on when the function stops and how it calls itself.

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