Find Peak Element Using Binary Search in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index for binary search.

DSA C++

int mid = low [1] (high - low) / 2;

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid = (low + high) / 2 which can cause overflow.

Using subtraction instead of addition.

We calculate mid as low plus half the distance between high and low: mid = low + (high - low) / 2.

2fill in blank

medium

Complete the condition to check if mid is a peak element.

DSA C++

if ((mid == 0 || nums[mid] > nums[mid - 1]) && (mid == n - 1 || nums[mid] > nums[[1]]))

Drag options to blanks, or click blank then click option'

Amid + 1

Bn - 1

Cmid

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid - 1 instead of mid + 1.

Comparing with mid instead of mid + 1.

We check if the element at mid is greater than the next element by comparing nums[mid] > nums[mid + 1].

3fill in blank

hard

Fix the error in updating the high pointer when the left neighbor is greater.

DSA C++

if (nums[mid] < nums[[1]]) {
    high = mid [2] 1;
}

Drag options to blanks, or click blank then click option'

Amid - 1

Bmid + 1

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid + 1 instead of mid - 1.

Using addition instead of subtraction.

If the left neighbor is greater, we move high to mid - 1 to search the left half.

4fill in blank

hard

Fill both blanks to update low pointer when right neighbor is greater and to return the peak index.

DSA C++

else {
    low = mid [1] 1;
}

return [2];

Drag options to blanks, or click blank then click option'

A+

B-

Cmid

Dlow

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition for low.

Returning low instead of mid.

When the right neighbor is greater, we move low to mid + 1. The peak index is mid when found.

5fill in blank

hard

Fill all three blanks to complete the binary search loop and return the peak element index.

DSA C++

while (low [1] high) {
    int mid = low + (high - low) / 2;
    if ((mid == 0 || nums[mid] > nums[mid - 1]) && (mid == n - 1 || nums[mid] > nums[mid + 1])) {
        return [2];
    } else if (mid > 0 && nums[mid] < nums[mid - 1]) {
        high = mid [3] 1;
    } else {
        low = mid + 1;
    }
}
return low;

Drag options to blanks, or click blank then click option'

A<=

Bmid

C-

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= in the loop condition.

Returning low instead of mid inside the loop.

Using addition instead of subtraction when updating high.

The loop continues while low is less than or equal to high. Return mid when peak found. Decrease high by 1 from mid when left neighbor is greater.

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