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

Find Peak Element Using Binary Search

📖 Scenario: You have a list of numbers representing heights of mountains in a range. A peak is a mountain that is taller than its neighbors. You want to find any one peak efficiently.

🎯 Goal: Build a Java program that uses binary search to find the index of a peak element in the list of mountain heights.

📋 What You'll Learn

Create an integer array called mountains with the exact values {1, 3, 20, 4, 1, 0}

Create an integer variable called left and set it to 0

Create an integer variable called right and set it to mountains.length - 1

Use a while loop with the condition left < right

Inside the loop, calculate mid as (left + right) / 2

Compare mountains[mid] and mountains[mid + 1] to decide which half to keep

Print the index of the peak element found

💡 Why This Matters

🌍 Real World

Finding a peak element is useful in signal processing, stock price analysis, and terrain mapping where local maximums are important.

💼 Career

Binary search and peak finding algorithms are common in software engineering interviews and real-world applications requiring efficient data analysis.

1

Create the mountain heights array

Create an integer array called mountains with these exact values: {1, 3, 20, 4, 1, 0}

DSA Java

// Create the mountains array
// Your code here

Hint

Use int[] mountains = {1, 3, 20, 4, 1, 0}; to create the array.

2

Set up binary search boundaries

Create two integer variables called left and right. Set left to 0 and right to mountains.length - 1

DSA Java

public class PeakFinder {
    public static void main(String[] args) {
        int[] mountains = {1, 3, 20, 4, 1, 0};
        // Create left and right variables
        // Your code here
    }
}

Hint

Use int left = 0; and int right = mountains.length - 1;.

3

Implement binary search to find a peak

Use a while loop with condition left < right. Inside the loop, calculate mid as (left + right) / 2. If mountains[mid] is less than mountains[mid + 1], set left to mid + 1. Otherwise, set right to mid

DSA Java

public class PeakFinder {
    public static void main(String[] args) {
        int[] mountains = {1, 3, 20, 4, 1, 0};
        int left = 0;
        int right = mountains.length - 1;
        // Add while loop to find peak
        // Your code here
    }
}

Hint

Use binary search logic comparing mountains[mid] and mountains[mid + 1] to move left or right.

4

Print the peak element index

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

DSA Java

public class PeakFinder {
    public static void main(String[] args) {
        int[] mountains = {1, 3, 20, 4, 1, 0};
        int left = 0;
        int right = mountains.length - 1;
        while (left < right) {
            int mid = (left + right) / 2;
            if (mountains[mid] < mountains[mid + 1]) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        // Print the peak index
        // Your code here
    }
}

Hint

Use System.out.println(left); to print the peak index.

Practice

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

easy

A. The smallest element in the array

B. An element that is greater than its neighbors

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: Relate to binary search usage

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

3. Final Answer:

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

4. Quick Check:

   Peak element = greater than neighbors [OK]

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

Common Mistakes:

• Confusing peak with minimum element
• Thinking peak must be at array ends only
• Assuming peak equals neighbors

2. Which of the following is the correct way to start the binary search for a peak element in Java?

easy

A. int left = 0, right = arr.length;

B. int left = 1, right = arr.length;

C. int left = 0, right = arr.length - 1;

D. int left = 1, right = arr.length - 1;

Solution

1. Step 1: Identify valid index range for binary search

   Array indices in Java go from 0 to length-1, so left should be 0 and right should be arr.length - 1.

2. Step 2: Check options for correct initialization

   int left = 0, right = arr.length - 1; correctly sets left=0 and right=arr.length-1, which covers the whole array.

3. Final Answer:

   int left = 0, right = arr.length - 1; -> Option C

4. Quick Check:

   Index range = 0 to length-1 [OK]

Hint: Array indices start at 0, end at length-1 [OK]

Common Mistakes:

• Starting right at arr.length causes out-of-bounds
• Starting left at 1 skips first element
• Using invalid index ranges

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

medium

A. 0

B. 1

C. 3

D. 2

Solution

1. Step 1: Identify peak elements in the array

   Array: [1, 3, 20, 4, 1, 0]. 20 at index 2 is greater than neighbors 3 and 4, so it's a peak.

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

   Binary search compares mid elements and moves left or right until it finds 20 at index 2.

3. Final Answer:

   2 -> Option D

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
• Ignoring neighbors comparison

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

int mid = left + (right - left) / 2;
if (arr[mid] < arr[mid + 1]) {
    left = mid + 1;
} else {
    right = mid - 1;
}
return left;

medium

A. It may cause ArrayIndexOutOfBoundsException at arr[mid + 1]

B. It returns right instead of left

C. It uses wrong mid calculation causing infinite loop

D. It should compare arr[mid] with arr[mid - 1]

Solution

1. Step 1: Check boundary conditions for arr[mid + 1]

   If mid is at the last index, arr[mid + 1] access causes ArrayIndexOutOfBoundsException.

2. Step 2: Identify missing boundary check

   The code does not check if mid + 1 is within array bounds before accessing.

3. Final Answer:

   It may cause ArrayIndexOutOfBoundsException at arr[mid + 1] -> Option A

4. Quick Check:

   Accessing arr[mid+1] at end causes error [OK]

Hint: Check array bounds before accessing mid+1 [OK]

Common Mistakes:

• Ignoring array boundary checks
• Assuming mid+1 always valid
• Confusing left and right updates

5. You have an array [10, 20, 15, 2, 23, 90, 67]. Using binary search to find a peak element, which index will the algorithm return?

hard

A. 5

B. 6

C. 4

D. 1

Solution

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

   Array: [10, 20, 15, 2, 23, 90, 67]. Peaks are 20 (index 1), 90 (index 5).

2. Step 2: Understand binary search behavior

   Binary search compares mid with neighbors and moves towards the side with higher neighbor. It will find 90 at index 5 as peak.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Peak found at index 5 = 90 [OK]

Hint: Binary search finds any peak, here index 5 with value 90 [OK]

Common Mistakes:

• Assuming only first peak is returned
• Confusing peak with global max
• Not tracing binary search steps carefully