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

Find Peak Element Using Binary Search

📖 Scenario: You are working with a list of numbers representing heights of hills along a trail. You want to find a hill that is taller than its neighbors, called a peak.Using a fast method called binary search, you will find one peak hill without checking every hill.

🎯 Goal: Build a Go program that finds a peak element in a list of integers using binary search.A peak element is one that is greater than its neighbors.

📋 What You'll Learn

Create a slice of integers called heights with the exact values: 1, 3, 20, 4, 1, 0

Create two integer variables left and right to hold the start and end indexes of the slice

Write a for loop that runs while left is less than right

Inside the loop, calculate the middle index mid and compare heights[mid] with heights[mid+1]

Adjust left or right based on the comparison to narrow down the peak location

After the loop ends, print the index of the peak element

💡 Why This Matters

🌍 Real World

Finding peak elements quickly is useful in signal processing, stock price analysis, and terrain mapping where you want to identify local maximum points.

💼 Career

Binary search and peak finding algorithms are common in software engineering interviews and are useful for optimizing search problems in real applications.

1

Create the heights slice

Create a slice of integers called heights with these exact values: 1, 3, 20, 4, 1, 0

DSA Go

# Create the slice heights with values 1, 3, 20, 4, 1, 0
# Your code here

Hint

Use heights := []int{1, 3, 20, 4, 1, 0} to create the slice.

2

Set left and right pointers

Create two integer variables called left and right. Set left to 0 and right to the last index of heights (which is 5).

DSA Go

package main

func main() {
    heights := []int{1, 3, 20, 4, 1, 0}
    // Create left and right variables
    // Your code here
}

Hint

Use left := 0 and right := len(heights) - 1.

3

Implement binary search loop to find peak

Write a for loop that runs while left < right. Inside the loop, calculate mid as (left + right) / 2. If heights[mid] is less than heights[mid+1], set left to mid + 1. Otherwise, set right to mid.

DSA Go

package main

func main() {
    heights := []int{1, 3, 20, 4, 1, 0}
    left := 0
    right := len(heights) - 1

    // Write a for loop to find the peak
    // Your code here
}

Hint

Use a for loop with condition left < right. Calculate mid inside the loop. Compare heights[mid] and heights[mid+1] to decide how to move left or right.

4

Print the peak element index

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

DSA Go

package main

import "fmt"

func main() {
    heights := []int{1, 3, 20, 4, 1, 0}
    left := 0
    right := len(heights) - 1

    for left < right {
        mid := (left + right) / 2
        if heights[mid] < heights[mid+1] {
            left = mid + 1
        } else {
            right = mid
        }
    }

    // Print the peak index
    // Your code here
}

Hint

Use fmt.Println(left) to print the peak index.