Find Peak Element Using Binary Search in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

function findPeakElement(nums) {
  let left = [1];
  let right = nums.length - 1;
  while (left < right) {
    // logic here
  }
  return left;
}

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

A0

B1

C-1

Dnums.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 or negative index causes skipping elements or errors.

The binary search starts from the first index 0 as the left boundary.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Javascript

while (left < right) {
  let mid = Math.floor((left [1] right) / 2);
  // logic here
}

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

A*

B-

C+

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition causes wrong mid calculation.

The middle index is calculated by adding left and right, then dividing by 2.

3fill in blank

hard

Fix the error in the condition to decide which side to search next.

DSA Javascript

if (nums[mid] [1] nums[mid + 1]) {
  left = mid + 1;
} else {
  right = mid;
}

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

A<

B>

C==

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses the search direction and breaks the logic.

If nums[mid] is less than nums[mid + 1], the peak is on the right side.

4fill in blank

hard

Fill both blanks to complete the binary search loop and return the peak index.

DSA Javascript

while ([1] < [2]) {
  let mid = Math.floor((left + right) / 2);
  if (nums[mid] < nums[mid + 1]) {
    left = mid + 1;
  } else {
    right = mid;
  }
}
return left;

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

Aleft

Bright

Cmid

Dnums.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or nums.length in the loop condition causes infinite loops or errors.

The loop continues while left is less than right to narrow down the peak.

5fill in blank

hard

Fill all three blanks to create a complete function that finds a peak element index.

DSA Javascript

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

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

Aleft

Bright

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up left and right variable names or using wrong comparison operators.

Initialize left and right pointers, then compare nums[mid] < nums[mid + 1] to decide direction.