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

Find Peak Element Using Binary Search

📖 Scenario: Imagine you have 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.

🎯 Goal: Build a program that finds a peak hill using a fast method called binary search.

📋 What You'll Learn

Create a vector called heights with exact values

Create two integer variables left and right for binary search boundaries

Use a while loop to apply binary search logic

Print the index of a peak element found

💡 Why This Matters

🌍 Real World

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

💼 Career

Binary search and peak finding are common in technical interviews and real-world software that needs fast data analysis.

1

Create the heights vector

Create a std::vector<int> called heights with these exact values: 1, 3, 20, 4, 1, 0

DSA C++

#include 
#include 

int main() {
    // Create the heights vector
    // Your code here
}

Hint

Use std::vector<int> heights = {1, 3, 20, 4, 1, 0}; to create the vector.

2

Set binary search boundaries

Create two integer variables called left and right. Set left to 0 and right to heights.size() - 1

DSA C++

#include 
#include 

int main() {
    std::vector heights = {1, 3, 20, 4, 1, 0};
    // Set left and right boundaries
    // Your code here
}

Hint

Use int left = 0; and int right = heights.size() - 1;.

3

Apply binary search to find a peak

Use a while loop with condition left < right. Inside the loop, create mid as left + (right - left) / 2. If heights[mid] < heights[mid + 1], set left = mid + 1, else set right = mid

DSA C++

#include 
#include 

int main() {
    std::vector heights = {1, 3, 20, 4, 1, 0};
    int left = 0;
    int right = heights.size() - 1;

    // Use binary search to find peak
    // Your code here
}

Hint

Use a while loop and update left or right based on comparing heights[mid] and heights[mid + 1].

4

Print the peak element index

Print the value of left using std::cout to show the index of a peak element

DSA C++

#include 
#include 

int main() {
    std::vector heights = {1, 3, 20, 4, 1, 0};
    int left = 0;
    int right = heights.size() - 1;

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

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

Hint

Use std::cout << left << std::endl; to print 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. An element that is smaller than its neighbors

C. The first element of the array

D. The last element of the array

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 only 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 search direction.

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 smaller [OK]

Common Mistakes:

• Confusing peak with smallest element
• Thinking peak is always first or last element
• Ignoring boundary conditions

2. Which of the following is the correct syntax to start binary search for peak element in C++?

easy

A. int left = 1, right = n;

B. int left = 0, right = n;

C. int left = 0, right = n - 1;

D. int left = 1, right = n - 1;

Solution

1. Step 1: Identify valid index range for array of size n

   Array indices go from 0 to n-1, so left should start at 0 and right at n-1.

2. Step 2: Check each option for correct initialization

   int left = 0, right = n - 1; correctly initializes left=0 and right=n-1, suitable for binary search.

3. Final Answer:

   int left = 0, right = n - 1; -> Option C

4. Quick Check:

   Index range = 0 to n-1 [OK]

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

Common Mistakes:

• Starting left at 1 instead of 0
• Setting right to n instead of n-1
• Confusing inclusive and exclusive 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. 2

C. 3

D. 1

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 B

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 as peak incorrectly
• Confusing index with value
• Not checking both neighbors

4. Given the following code snippet, what is the bug that prevents finding the peak element correctly?

int findPeak(int arr[], int n) {
  int left = 0, right = n - 1;
  while (left < right) {
    int mid = (left + right) / 2;
    if (arr[mid] < arr[mid + 1])
      left = mid + 1;
    else
      right = mid - 1;
  }
  return left;
}

medium

A. Decreasing right by 1 skips potential peak

B. Mid calculation causes overflow

C. Loop condition should be left <= right

D. Returning left instead of right

Solution

1. Step 1: Analyze the binary search update steps

   If arr[mid] >= arr[mid+1], the code sets right = mid - 1, which skips mid as a candidate peak.

2. Step 2: Correct approach is to set right = mid to include mid in search

   Setting right = mid keeps mid in range; subtracting 1 may miss the peak.

3. Final Answer:

   Decreasing right by 1 skips potential peak -> Option A

4. Quick Check:

   Right update must keep mid included [OK]

Hint: Do not exclude mid when moving right pointer [OK]

Common Mistakes:

• Setting right = mid - 1 instead of right = mid
• Ignoring mid element as possible peak
• Incorrect loop condition

5. You have an array [10, 20, 15, 2, 23, 90, 67]. Using binary search for peak element, which index will be returned and why?

hard

A. 0, because 10 is the first element

B. 1, because 20 is a peak element

C. 6, because 67 is the last element

D. 5, because 90 is a peak element

Solution

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

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

2. Step 2: Binary search tends to find a peak, often the higher one

   Binary search will find 90 at index 5 because it is a higher peak than 20 at index 1.

3. Final Answer:

   5, because 90 is a peak element -> Option D

4. Quick Check:

   Highest peak found = index 5 [OK]

Hint: Binary search finds any peak; higher peaks likely found first [OK]

Common Mistakes:

• Assuming first peak is always returned
• Ignoring multiple peaks in array
• Confusing last element as peak without checking neighbors