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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the mid index in 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 '-' instead of '+' causes mid to be less than low.

Using '*' or '/' directly without proper parentheses causes wrong calculation.

The mid index is calculated by adding half the distance between low and high to low, so '+' is correct.

2fill in blank

medium

Complete the condition to check if mid element is less than its next element.

DSA C

if (nums[mid] [1] nums[mid + 1]) {

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

A>

B==

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' reverses the logic and leads to wrong search direction.

Using '==' or '!=' does not help decide the search direction.

If nums[mid] is less than nums[mid + 1], we move right to find the peak.

3fill in blank

hard

Fix the error in updating the high index when moving left.

DSA C

else {
    high = [1];
}

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

Amid - 1

Bmid + 1

Clow + 1

Dhigh - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting high to mid + 1 causes infinite loop or wrong search.

Using low + 1 or high - 1 unrelated to mid causes incorrect boundaries.

When moving left, high should be set to mid - 1 to exclude mid from next search.

4fill in blank

hard

Fill both blanks to complete the binary search loop and return statement.

DSA C

while (low [1] high) {
    int mid = low + (high - low) / 2;
    if (nums[mid] < nums[mid + 1]) {
        low = mid + 1;
    } else {
        high = mid - 1;
    }
}
return [2];

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

A<=

B<

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' in loop condition may miss the peak at the boundary.

Returning high instead of low may give wrong index.

The loop runs while low is less than or equal to high, and the peak index is at low after the loop.

5fill in blank

hard

Fill all three blanks to complete the function signature, variable declaration, and return statement.

DSA C

[1] findPeakElement([2] nums[], int numsSize) {
    int low = 0, high = numsSize - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (nums[mid] < nums[mid + 1]) {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    return [3];
}

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

Aint

Clow

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using void as return type causes compilation error.

Returning high instead of low may give wrong peak index.

The function returns an int index, takes an int array, and returns low as the peak index.

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