Why Binary Search and What Sorted Order Gives You in DSA Javascript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start binary search with the correct initial left index.

DSA Javascript

let left = [1];

Drag options to blanks, or click blank then click option'

A1

B0

C-1

Darray.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 or negative index causes skipping the first element.

The left index in binary search starts at 0, the first position of the sorted array.

2fill in blank

medium

Complete the code to calculate the middle index correctly in binary search.

DSA Javascript

let mid = Math.floor((left + [1]) / 2);

Drag options to blanks, or click blank then click option'

Aright

Barray.length

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using array.length instead of right causes wrong mid calculation.

The middle index is the floor of the average of left and right indices.

3fill in blank

hard

Fix the error in updating the left index when the middle element is less than the target.

DSA Javascript

if (array[mid] < target) {
  left = [1];
}

Drag options to blanks, or click blank then click option'

Amid

Bright

Cleft + 1

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting left to mid causes infinite loop if mid doesn't change.

When middle element is less than target, search moves right, so left becomes mid + 1.

4fill in blank

hard

Fill both blanks to update right index and return found index correctly.

DSA Javascript

if (array[mid] === target) {
  return [1];
} else if (array[mid] > target) {
  [2] = mid - 1;
}

Drag options to blanks, or click blank then click option'

Amid

Bleft

Cright

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Returning target instead of index; updating wrong boundary variable.

Return mid when target found; update right to mid - 1 when middle is greater than target.

5fill in blank

hard

Fill all three blanks to complete the binary search function.

DSA Javascript

function binarySearch(array, target) {
  let left = 0;
  let right = array.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (array[mid] === target) {
      return [1];
    } else if (array[mid] < target) {
      left = [2];
    } else {
      right = [3];
    }
  }
  return -1;
}

Drag options to blanks, or click blank then click option'

Amid

Bmid + 1

Cmid - 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Incorrect boundary updates cause infinite loops or missed targets.

Return mid if found; move left to mid + 1 if middle less than target; move right to mid - 1 if middle greater.

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