Find Peak Element Using Binary Search in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a peak element in an array?

A peak element is an element that is greater than its neighbors. For the first or last element, only one neighbor is considered.

intermediate

Why is binary search suitable for finding a peak element?

Because if an element is not a peak, we can decide which side to move based on the neighbor's value, reducing the search space by half each time.

beginner

In the binary search for peak element, if mid element is less than its right neighbor, which side do we search next?

We search the right half because a peak must exist there.

beginner

What is the time complexity of finding a peak element using binary search?

O(log n), where n is the number of elements in the array.

beginner

Show the base condition to stop the binary search when finding a peak element.

When the search space reduces to one element, that element is a peak.

What defines a peak element in an array?

A An element smaller than its neighbors

B The first element only

C An element greater than its neighbors

D The last element only

A peak element is one that is greater than its neighbors.

If mid element is less than its left neighbor, where should binary search continue?

A Left half

B Either half

C Right half

D Stop search

Search left half because a peak exists there.

What is the worst-case time complexity of binary search for peak element?

A O(log n)

B O(n)

C O(n log n)

D O(1)

Binary search reduces search space by half each step, so O(log n).

Can the first or last element be a peak element?

A Only the last element can be

B No, never

C Only the first element can be

D Yes, if greater than its only neighbor

First or last element can be peak if greater than their single neighbor.

What happens if mid element is greater than both neighbors?

A Search left half

B Mid is a peak element

C Search right half

D Continue searching both halves

If mid is greater than neighbors, it is a peak.

Explain how binary search helps find a peak element in an array.

Think about how neighbors guide the search direction.

Describe the steps and conditions to implement binary search for finding a peak element.

Focus on the loop and decision making.

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