Find Peak Element Using Binary Search in DSA C - Time & Space Complexity

Time Complexity: Find Peak Element Using Binary Search

O(log n)

Understanding Time Complexity

We want to understand how fast the binary search method finds a peak element in an array.

The question is: how does the time taken grow as the array size grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int findPeakElement(int* nums, int numsSize) {
    int left = 0, right = numsSize - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[mid + 1]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}
    

This code finds a peak element index in the array by repeatedly narrowing the search range using binary search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that halves the search range each time.
• How many times: The loop runs about log base 2 of the array size times.

How Execution Grows With Input

Each step cuts the search area roughly in half, so the number of steps grows slowly as the array gets bigger.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: Doubling the input size adds only one more step, showing slow growth.

Final Time Complexity

Time Complexity: O(log n)

This means the time to find a peak grows slowly, increasing by one step each time the input size doubles.

Common Mistake

[X] Wrong: "The algorithm checks every element, so it must be O(n)."

[OK] Correct: The code does not check every element; it smartly halves the search space each time, so it runs much faster than checking all elements.

Interview Connect

Understanding this binary search approach shows you can use smart searching to solve problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if we changed the array to be sorted strictly increasing or strictly decreasing? How would the time complexity change?"

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