Binary Search as Divide and Conquer in DSA Javascript - 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 in half, then focus on the half where the target value could be, repeating this until the target is found or the search space is empty.

beginner

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

If the target value is less than the middle element, we search the left half because the array is sorted and the target can only be there.

beginner

Why must the array be sorted for binary search to work?

Because binary search relies on comparing the target with the middle element to decide which half to search next, sorting ensures this decision is correct.

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

How does binary search demonstrate the divide and conquer strategy?

It divides the array into two halves and conquers by recursively searching the half that may contain the target, combining results to find the answer.

What is the first step in binary search?

A Find the middle element of the array

B Sort the array

C Check the first element

D Check the last element

Binary search starts by finding the middle element to compare with the target.

If the target is greater than the middle element, which part of the array do we search next?

A None, target not found

B Left half

C Whole array again

D Right half

If the target is greater, we search the right half because the array is sorted.

What happens if the target is equal to the middle element?

A Search right half

B Return the middle element's index

C Search left half

D Continue searching both halves

If the target equals the middle element, we have found it and return its index.

Which of these is NOT a requirement for binary search?

A Array must contain unique elements

B Array must be sorted

C Array must be divided into halves

D Target must be comparable

Binary search works even if elements repeat; uniqueness is not required.

What is the worst-case time complexity of binary search?

A O(n)

B O(1)

C O(log n)

D O(n log n)

Binary search reduces the search space by half each time, leading to O(log n) complexity.

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

Think about how the array is split and how the search space shrinks.

Describe why binary search is more efficient than linear search for large sorted arrays.

Focus on how quickly the search space reduces.

Practice

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

easy

A. It sorts the array before searching for the target.

B. It checks every element one by one until it finds the target.

C. It repeatedly divides the search interval in half to find the target value.

D. It uses random guesses to find the target quickly.

Solution

1. Step 1: Understand divide and conquer in binary search

   Binary search splits the array into two halves and decides which half to search next.

2. Step 2: Identify the repeated halving process

   It keeps dividing the search space by half until the target is found or no elements remain.

3. Final Answer:

   It repeatedly divides the search interval in half to find the target value. -> Option C

4. Quick Check:

   Divide and conquer = halve search space [OK]

Hint: Binary search halves the array each step to find the target [OK]

Common Mistakes:

• Thinking binary search checks every element
• Confusing binary search with linear search
• Assuming binary search sorts the array

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

easy

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

B. const mid = low + high / 2;

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

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

Solution

1. Step 1: Understand middle index calculation

   The middle index should be an integer, so we use Math.floor to round down.

2. Step 2: Check each option

   const mid = (low + high) / 2; returns a float, const mid = Math.floor((low + high) / 2); correctly floors the average, const mid = Math.ceil((low + high) / 2); rounds up which can cause errors, const mid = low + high / 2; has wrong operator precedence.

3. Final Answer:

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

4. Quick Check:

   Use Math.floor to get integer mid [OK]

Hint: Use Math.floor((low + high) / 2) for middle index [OK]

Common Mistakes:

• Not flooring the middle index causing float errors
• Using wrong operator precedence without parentheses
• Using Math.ceil which can skip elements

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

function binarySearch(arr, target) {
  let low = 0;
  let high = arr.length - 1;
  while (low <= high) {
    const 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;
}

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

medium

A. 2

B. 3

C. -1

D. 4

Solution

1. Step 1: Trace the binary search for target 6

   Initial low=0, high=4, mid=2 (element 6). arr[mid] equals target, so return 2.

2. Step 2: Confirm returned index

   The function returns index 2 where 6 is found.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Check mid index value matches target to return index [OK]

Common Mistakes:

• Confusing index with value
• Off-by-one errors in mid calculation
• Returning value instead of index

4. Identify the bug in this binary search code snippet and choose the correct fix:

function binarySearch(arr, target) {
  let low = 0;
  let high = arr.length;
  while (low <= high) {
    const 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. Change while condition to low < high

B. Change high initialization to arr.length - 1

C. Use Math.ceil instead of Math.floor for mid

D. Return arr[mid] instead of mid

Solution

1. Step 1: Check high initialization

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

2. Step 2: Confirm fix

   Changing high to arr.length - 1 prevents accessing undefined elements and infinite loop.

3. Final Answer:

   Change high initialization to arr.length - 1 -> Option B

4. Quick Check:

   High = arr.length - 1 fixes index error [OK]

Hint: Set high to last index: arr.length - 1 [OK]

Common Mistakes:

• Setting high to arr.length causing out-of-bounds
• Changing while condition incorrectly
• Returning value instead of index

5. You have a sorted array arr = [1, 3, 3, 5, 7, 7, 7, 9]. Using binary search as divide and conquer, which index will be returned when searching for 7 with the standard binary search implementation below?

function binarySearch(arr, target) {
  let low = 0;
  let high = arr.length - 1;
  while (low <= high) {
    const 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;
}

Choose the correct index returned and explain why.

hard

A. 5 (middle occurrence of 7)

B. 6 (last occurrence of 7)

C. 4 (first occurrence of 7)

D. -1 (not found)

Solution

1. Step 1: Trace binary search on duplicates

   Initial low=0, high=7, mid=3 (arr[3]=5 < 7), so low=4. Next mid=5 (arr[5]=7), matches target, returns 5 immediately.

2. Step 2: Understand standard binary search behavior

   Standard binary search returns the first found match, not necessarily first or last occurrence.

3. Final Answer:

   5 (middle occurrence of 7) -> Option A

4. Quick Check:

   Standard binary search returns any matching index [OK]

Hint: Standard binary search returns any matching index, not first or last [OK]

Common Mistakes:

• Assuming it always returns first or last occurrence
• Ignoring duplicates in array
• Expecting -1 when target exists