Find Peak Element Using Binary Search in DSA Java - 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 useful for finding a peak element?

Binary search helps find a peak element efficiently by checking the middle element and deciding which half to search next, reducing time from linear to logarithmic.

beginner

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

We search the right half because a peak must exist there since the right neighbor is greater.

beginner

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

The time complexity is O(log n) because the search space is halved in each step.

intermediate

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

When the middle element is greater than or equal to its neighbors, it is a peak, so we return its index.

What defines a peak element in an array?

A An element smaller than its neighbors

B An element greater than its neighbors

C The first element only

D The last element only

A peak element is greater than its neighbors, including edge cases where only one neighbor exists.

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

A Left half

B Search both halves

C Return mid as peak

D Right half

If mid is less than left neighbor, a peak must exist on the left side.

What is the time complexity of the binary search approach to find a peak element?

A O(n)

B O(n log n)

C O(log n)

D O(1)

Binary search reduces the search space by half each time, so time complexity is O(log n).

What happens if the middle element is greater than both neighbors?

A Search right half

B Search left half

C Continue searching both sides

D It is a peak element

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

Which of these is NOT a valid step in binary search for peak element?

A Always move to left half

B Move to half where neighbor is greater

C Compare mid with neighbors

D Return mid if it is peak

We do not always move left; direction depends on which neighbor is greater.

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

Think about how you reduce the search space by half each time.

Describe the conditions to identify a peak element during binary search.

Focus on comparisons between mid and neighbors.

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