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 -- a peak hill. This helps hikers find the highest points easily.

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

📋 What You'll Learn

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

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

Implement a binary search loop using while (left < right) to find a peak element index

Print the index of the peak element found

💡 Why This Matters

🌍 Real World

Finding peak elements helps in signal processing, stock price analysis, and identifying local maxima in data.

💼 Career

Binary search and peak finding are common algorithmic problems asked in technical interviews and used in performance-critical applications.

1

Create the array of hill heights

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

DSA C

# Create the array nums with values {1, 3, 20, 4, 1, 0}
// Your code here

Hint

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

2

Set up the search boundaries

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

DSA C

int nums[] = {1, 3, 20, 4, 1, 0};
// Create int variables left = 0 and right = 5
// Your code here

Hint

Remember the array length is 6, so last index is 5.

3

Implement binary search to find peak

Use a while (left < right) loop. Inside, calculate mid = (left + right) / 2. If nums[mid] is less than nums[mid + 1], set left = mid + 1. Otherwise, set right = mid.

DSA C

int nums[] = {1, 3, 20, 4, 1, 0};
int left = 0;
int right = 5;
// Use while loop to find peak index
// Your code here

Hint

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

4

Print the peak element index

Print the value of left which holds the index of the peak element using printf.

DSA C

int nums[] = {1, 3, 20, 4, 1, 0};
int left = 0;
int right = 5;
while (left < right) {
    int mid = (left + right) / 2;
    if (nums[mid] < nums[mid + 1]) {
        left = mid + 1;
    } else {
        right = mid;
    }
}
// Print the peak index stored in left
// Your code here

Hint

The peak element index is stored in left. Use printf("%d\n", 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 the 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 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 syntax to calculate the middle index in binary search in C?

easy

A. int mid = (high - low) * 2;

B. int mid = (low + high) / 2;

C. int mid = (low - high) / 2;

D. int mid = low + (high - low) / 2;

Solution

1. Step 1: Understand integer overflow risk in mid calculation

   Using (low + high) / 2 can overflow if low and high are large integers.

2. Step 2: Use safer formula to avoid overflow

   Calculate mid as low + (high - low) / 2 to prevent overflow.

3. Final Answer:

   int mid = low + (high - low) / 2; -> Option D

4. Quick Check:

   Safe mid calculation = low + (high - low) / 2 [OK]

Hint: Use low + (high - low)/2 to avoid overflow [OK]

Common Mistakes:

• Using (low + high)/2 causing overflow
• Subtracting incorrectly in mid calculation
• Multiplying instead of dividing

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

medium

A. 1

B. 2

C. 3

D. 0

Solution

1. Step 1: Identify peak candidates in the array

   Array: [1, 3, 20, 4, 1, 0]. Element 20 at index 2 is greater than neighbors 3 and 4.

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

   Binary search compares mid elements and moves towards the peak, landing on index 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Peak element index = 2 [OK]

Hint: Peak is local max, here 20 at index 2 [OK]

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing index with value
• Ignoring neighbors in peak definition

4. Consider this code snippet to find a peak element index:

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

What is the bug in this code?

medium

A. It should return high instead of low

B. The while condition should be low <= high

C. The else branch should set high = mid, not mid - 1

D. The mid calculation causes overflow

Solution

1. Step 1: Analyze the else branch update of high

   Setting high = mid - 1 skips the mid element which could be a peak.

2. Step 2: Correct update to include mid in search

   Change to high = mid to keep mid in the search space for peak.

3. Final Answer:

   The else branch should set high = mid, not mid - 1 -> Option C

4. Quick Check:

   Keep mid in search by high = mid [OK]

Hint: Use high = mid to avoid skipping potential peak [OK]

Common Mistakes:

• Reducing high too early and missing peak
• Changing while condition unnecessarily
• Ignoring mid element in binary search

5. Given an array [10, 20, 15, 2, 23, 90, 67], how many peak elements can binary search find and which index will it return if it always moves towards the right when equal?

hard

A. 2 peaks, returns index 5

B. 3 peaks, returns index 1

C. 1 peak, returns index 6

D. 2 peaks, returns index 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) and 90 (index 5) as both are greater than neighbors.

2. Step 2: Understand binary search behavior moving right on equal

   Binary search will move towards the right peak at index 5 because it moves right when equal or increasing.

3. Final Answer:

   2 peaks, returns index 5 -> Option A

4. Quick Check:

   Two peaks, rightmost peak index 5 [OK]

Hint: Multiple peaks possible; binary search returns one found by direction [OK]

Common Mistakes:

• Assuming only one peak always
• Ignoring binary search direction rules
• Confusing peak count with returned index