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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index for binary search.

DSA Typescript

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

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

Aceil

Bfloor

Cround

Dtrunc

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.ceil causes index to go out of range sometimes.

Using Math.round can cause uneven splits.

We use Math.floor to get the middle index as an integer by rounding down.

2fill in blank

medium

Complete the condition to check if mid element is a peak.

DSA Typescript

if ((mid === 0 || nums[mid] > nums[mid - 1]) && (mid === nums.length - 1 || nums[mid] > nums[[1]])) {

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

Anums.length

Bmid - 1

Cmid

Dmid + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 instead of mid + 1 causes wrong comparison.

Using nums.length as index causes out of bounds error.

We compare the mid element with the next element at mid + 1 to check if it is greater.

3fill in blank

hard

Fix the error in updating the left pointer when mid element is less than the next element.

DSA Typescript

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

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

Amid + 1

Bmid

Cleft + 1

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using right - 1 is wrong direction.

We move left to mid + 1 to search the right half where a peak may exist.

4fill in blank

hard

Fill both blanks to complete the binary search loop condition and return statement.

DSA Typescript

while ([1] < [2]) {
    // binary search steps
}
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 in loop condition causes errors.

Returning right instead of left can be confusing but works; here we use left.

The loop runs while left < right. After the loop, left is the peak index.

5fill in blank

hard

Fill all three blanks to complete the function signature, initialize pointers, and return the peak index.

DSA Typescript

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

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

Anums

Bleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter name causes errors.

Returning right instead of left is possible but inconsistent here.

The function takes nums array, initializes left pointer, and returns left as the peak index.

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 just one side if at the boundary).

2. Step 2: Apply this to binary search context

   Binary search finds such an element by comparing mid element with neighbors to decide which half to search next.

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 [OK]

Common Mistakes:

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

2. Which of the following is the correct TypeScript syntax to start a binary search loop for finding a peak element in array nums?

easy

A. while(left < right) { ... }

B. if(nums[mid] > nums[mid+1]) { ... }

C. for(let i = 0; i < nums.length; i++)

D. switch(nums[mid]) { ... }

Solution

1. Step 1: Identify the loop type for binary search

   Binary search uses a while loop with condition left < right to narrow down the search range.

2. Step 2: Check options for correct binary search syntax

   while(left < right) { ... } shows the correct while loop structure for binary search.

3. Final Answer:

   while(left < right) { ... } -> Option A

4. Quick Check:

   Binary search uses while loop [OK]

Hint: Binary search uses while(left < right) loop [OK]

Common Mistakes:

• Using for loop instead of while
• Using if instead of loop for search
• Using switch incorrectly for binary search

3. What will be the output index of a peak element for the array nums = [1, 3, 20, 4, 1, 0] using binary search peak finding?

medium

A. 3

B. 1

C. 2

D. 0

Solution

1. Step 1: Identify peak elements in the array

   Check neighbors: 20 at index 2 is greater than 3 and 4, so it's a peak.

2. Step 2: Binary search will find index 2 as peak

   Binary search compares mid with neighbors and narrows to index 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Peak at index 2 = 20 [OK]

Hint: Peak is local max, here 20 at index 2 [OK]

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing peak with global max always
• Ignoring neighbors in peak definition

4. Given this TypeScript code snippet to find a peak element, what is the bug?

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

medium

A. Return should be right instead of left

B. The loop condition should be while(left <= right)

C. The mid calculation is incorrect

D. The else branch should set right = mid, not mid - 1

Solution

1. Step 1: Analyze the else branch update of right pointer

   Setting right = mid - 1 skips the mid element, possibly missing peak.

2. Step 2: Correct approach is right = mid to keep mid in range

   To ensure peak is not skipped, right should be set to mid, not mid - 1.

3. Final Answer:

   The else branch should set right = mid, not mid - 1 -> Option D

4. Quick Check:

   Keep mid in range by right = mid [OK]

Hint: Use right = mid to avoid skipping mid element [OK]

Common Mistakes:

• Setting right = mid - 1 causing missed peaks
• Changing loop condition unnecessarily
• Miscomputing mid index

5. You have an array nums = [10, 20, 15, 2, 23, 90, 67]. Using binary search to find a peak element, which index could be returned as a valid peak?

hard

A. 2

B. 5

C. 6

D. 3

Solution

1. Step 1: Identify all peak elements in the array

   Peaks are elements greater than neighbors: 20 at index 1 and 90 at index 5.

2. Step 2: Binary search can return any peak; index 5 (value 90) is a valid peak

   90 is greater than 23 and 67, so index 5 is a valid peak.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   90 at index 5 is peak [OK]

Hint: Binary search may return any peak, here index 5 is valid [OK]

Common Mistakes:

• Assuming only one peak exists
• Choosing non-peak elements
• Ignoring neighbors when checking peak