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.

Practice

1. What is a peak element in an array when using binary search to find it?

easy

A. An element that is greater than its neighbors

B. The smallest element in the array

C. The first element of the array

D. An element equal to its neighbors

Solution

1. Step 1: Understand the definition of a peak element

   A peak element is one that is strictly greater than its neighbors on both sides (or one side if at the boundary).

2. Step 2: Compare options with the definition

   Only An element that is greater than its neighbors matches this definition exactly.

3. Final Answer:

   An element that is greater than its neighbors -> Option A

4. Quick Check:

   Peak element = greater than neighbors [OK]

Hint: Peak element is bigger than neighbors, not smallest or equal [OK]

Common Mistakes:

• Thinking peak means smallest element
• Confusing peak with first or last element
• Assuming peak equals neighbors

2. Which of the following is the correct way to calculate the middle index mid in binary search in JavaScript?

easy

A. const mid = (left + right) / 2;

B. const mid = Math.ceil((left + right) / 2);

C. const mid = Math.floor((left + right) / 2);

D. const mid = left + right / 2;

Solution

1. Step 1: Understand integer division in JavaScript

   JavaScript division returns a float, so we use Math.floor to get an integer index.

2. Step 2: Check each option

   const mid = Math.floor((left + right) / 2); uses Math.floor on the average of left and right, which is correct for binary search.

3. Final Answer:

   const mid = Math.floor((left + right) / 2); -> Option C

4. Quick Check:

   Use Math.floor for integer mid index [OK]

Hint: Use Math.floor((left + right) / 2) for mid index [OK]

Common Mistakes:

• Not using Math.floor causes float index
• Using only division without parentheses
• Incorrect operator precedence in calculation

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

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

console.log(findPeak([1, 3, 2, 4, 1]));

medium

A. 3

B. 1

C. 2

D. 0

Solution

1. Step 1: Trace the binary search steps on array [1,3,2,4,1]

   Initial left=0, right=4. mid=2 (value=2). Compare nums[2]=2 and nums[3]=4. Since 2 < 4, left=mid+1=3.

2. Step 2: Next iteration with left=3, right=4

   mid=3 (value=4). Compare nums[3]=4 and nums[4]=1. Since 4 > 1, right=mid=3.

3. Step 3: Loop ends as left=right=3

   Return left=3, which is index of peak element 4.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Peak index = 3 [OK]

Hint: Compare mid and mid+1 to decide left or right [OK]

Common Mistakes:

• Returning value instead of index
• Not updating left or right correctly
• Confusing mid calculation

4. Identify the bug in this binary search code to find a peak element:

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

medium

A. The comparison should be nums[mid] > nums[mid + 1] instead of <.

B. The mid calculation is incorrect.

C. The return statement should return right instead of left.

D. The loop condition should be left < right instead of <=.

Solution

1. Step 1: Analyze the loop condition

   The loop uses left <= right, which can cause mid+1 to go out of bounds when mid = right.

2. Step 2: Correct loop condition for safe mid+1 access

   Changing to left < right ensures mid+1 is always valid inside the loop.

3. Final Answer:

   The loop condition should be left < right instead of <=. -> Option D

4. Quick Check:

   Loop condition left < right avoids out-of-bounds [OK]

Hint: Use left < right to avoid mid+1 overflow [OK]

Common Mistakes:

• Using <= causes index out of range error
• Changing mid calculation unnecessarily
• Returning wrong index

5. Given a strictly increasing then strictly decreasing array nums, how can you modify the binary search to find the peak element efficiently?
Choose the best approach.

hard

A. Use binary search comparing nums[mid] and nums[mid-1], move left or right accordingly.

B. Use binary search comparing nums[mid] and nums[mid+1], move left or right accordingly.

C. Scan the array linearly from start to find the maximum element.

D. Sort the array first, then pick the middle element.

Solution

1. Step 1: Understand the array shape

   The array increases then decreases, so the peak is the maximum element where nums[mid] > nums[mid+1].

2. Step 2: Use binary search comparing nums[mid] and nums[mid+1]

   If nums[mid] > nums[mid+1], move right pointer to mid; else move left pointer to mid+1.

3. Step 3: Avoid linear scan or sorting

   Linear scan is O(n), sorting changes order and is unnecessary.

4. Final Answer:

   Use binary search comparing nums[mid] and nums[mid+1], move left or right accordingly. -> Option B

5. Quick Check:

   Compare mid and mid+1 for peak in bitonic array [OK]

Hint: Compare mid and mid+1 to find peak in bitonic array [OK]

Common Mistakes:

• Using linear scan instead of binary search
• Sorting array destroys peak property
• Comparing mid and mid-1 incorrectly