Find Peak Element Using Binary Search in DSA Typescript - Build from Scratch

Find Peak Element Using Binary Search

📖 Scenario: Imagine you have a list of numbers representing the heights of hills along a trail. You want to find a hill that is taller than its neighbors, called a peak. This helps hikers find the highest point nearby quickly.

🎯 Goal: Build a program that uses binary search to find any peak element in a list of numbers efficiently.

📋 What You'll Learn

Create an array called heights with specific numbers

Create a variable left and right to hold the start and end indexes

Use a while loop with binary search logic to find a peak element index

Print the index of the peak element found

💡 Why This Matters

🌍 Real World

Finding peak elements is useful in signal processing, stock price analysis, and terrain mapping where local maximum points matter.

💼 Career

Understanding binary search and peak finding algorithms is important for software engineers working on performance-critical applications and data analysis.

1

Create the heights array

Create an array called heights with these exact numbers: [1, 3, 20, 4, 1, 0]

DSA Typescript

// Create the array heights with the given numbers
// Your code here

Hint

Use const heights = [1, 3, 20, 4, 1, 0]; to create the array.

2

Set up binary search boundaries

Create two variables called left and right. Set left to 0 and right to the last index of heights.

DSA Typescript

const heights = [1, 3, 20, 4, 1, 0];
// Create left = 0 and right = heights.length - 1
// Your code here

Hint

Use let left = 0; and let right = heights.length - 1;.

3

Implement binary search to find peak

Use a while loop with the condition left < right. Inside the loop, calculate mid as the middle index between left and right. If heights[mid] is less than heights[mid + 1], set left to mid + 1. Otherwise, set right to mid.

DSA Typescript

const heights = [1, 3, 20, 4, 1, 0];
let left = 0;
let right = heights.length - 1;
// Add a while loop with condition left < right
// Calculate mid inside the loop
// Update left or right based on comparison
// Your code here

Hint

Use Math.floor((left + right) / 2) to find mid. Compare heights[mid] and heights[mid + 1] to decide which side to keep.

4

Print the peak element index

Print the value of left, which is the index of a peak element.

DSA Typescript

const heights = [1, 3, 20, 4, 1, 0];
let left = 0;
let right = heights.length - 1;

while (left < right) {
    const mid = Math.floor((left + right) / 2);
    if (heights[mid] < heights[mid + 1]) {
        left = mid + 1;
    } else {
        right = mid;
    }
}
// Print the peak index left
// Your code here

Hint

The peak element index is stored in left. Use console.log(left); to print it.

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