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.