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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start binary search with the correct left index.

DSA Rust

let mut left = [1];

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

A1

B0

C-1

Dnums.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 skips the first element.

Using negative index is invalid in Rust.

Setting left to nums.len() is out of bounds.

The binary search starts from the first index, which is 0.

2fill in blank

medium

Complete the code to calculate the middle index safely.

DSA Rust

let mid = left + (right - left) [1] 2;

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

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division.

Adding instead of dividing.

Subtracting instead of dividing.

We divide the difference by 2 to find the middle index.

3fill in blank

hard

Fix the error in the condition to check if mid element is less than its right neighbor.

DSA Rust

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 greater than reverses the logic.

Using equality does not help find peak.

Using not equal is incorrect here.

We check if the middle element is less than the next element to decide search direction.

4fill in blank

hard

Fill both blanks to update the search boundaries correctly.

DSA Rust

if nums[mid] [1] nums[mid + 1] {
    left = mid [2] 1;
} else {
    right = mid;
}

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

A<

B+

C-

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < changes logic.

Subtracting 1 moves left boundary backward incorrectly.

If mid element is less than next, move left to mid + 1 to search right half.

5fill in blank

hard

Fill all three blanks to return the peak index after binary search loop.

DSA Rust

while left < right {
    let mid = left + (right - left) [1] 2;
    if nums[mid] [2] nums[mid + 1] {
        left = mid [3] 1;
    } else {
        right = mid;
    }
}
return left;

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

A/

B<

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of division for mid.

Using > instead of < in condition.

Using -1 instead of +1 to update left.

Divide by 2 to find mid, compare with <, and move left to mid + 1 if condition true.

Practice

1. What is a peak element in an array when using binary search to find it?

easy

A. An element equal to its neighbors

B. The smallest element in the array

C. The first element of the array

D. An element greater than 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 an edge).

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 greater than its neighbors -> Option D

4. Quick Check:

   Peak element = element greater than neighbors [OK]

Hint: Peak element is bigger than neighbors, not smallest [OK]

Common Mistakes:

• Confusing peak with smallest element
• Thinking peak must be at array start
• Assuming peak equals neighbors

2. Which Rust code snippet correctly initializes the binary search boundaries for finding a peak element in a vector nums?

easy

A. let (mut left, mut right) = (0, nums.len() - 1);

B. let (left, right) = (1, nums.len());

C. let (mut left, mut right) = (1, nums.len() - 1);

D. let (left, right) = (0, nums.len());

Solution

1. Step 1: Identify correct boundary initialization

   Binary search on array indices starts at 0 and ends at length - 1 because indexing is zero-based.

2. Step 2: Check mutability and correct tuple assignment

   We need mutable variables for left and right to update during search, so use mut and assign (0, nums.len() - 1).

3. Final Answer:

   let (mut left, mut right) = (0, nums.len() - 1); -> Option A

4. Quick Check:

   Start=0, End=length-1, mutable [OK]

Hint: Start at 0, end at length-1 with mut for binary search [OK]

Common Mistakes:

• Using length instead of length - 1 as right boundary
• Not declaring variables as mutable
• Starting from 1 instead of 0

3. What will be the output index of the peak element for the input array nums = vec![1, 3, 2, 1] using binary search peak finding?

medium

A. 0

B. 1

C. 2

D. 3

Solution

1. Step 1: Analyze the array and identify peak candidates

   Array is [1, 3, 2, 1]. Element at index 1 is 3, which is greater than neighbors 1 and 2.

2. Step 2: Confirm binary search would find index 1

   Binary search compares mid with neighbors and moves left or right. Mid initially at index 1 or 2 will find 3 as peak.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Peak element 3 at index 1 [OK]

Hint: Peak is local max, here 3 at index 1 [OK]

Common Mistakes:

• Choosing first or last index incorrectly
• Confusing peak with any max element
• Ignoring neighbors comparison

4. Identify the bug in this Rust code snippet for finding a peak element using binary search:

let mut left = 0;
let mut right = nums.len();
while left < right {
    let mid = (left + right) / 2;
    if nums[mid] < nums[mid + 1] {
        left = mid + 1;
    } else {
        right = mid;
    }
}
return left;

medium

A. Right boundary should be nums.len() - 1

B. Mid calculation causes overflow

C. Comparison should be nums[mid] > nums[mid + 1]

D. Loop condition should be left <= right

Solution

1. Step 1: Check boundary initialization

   Right boundary is set to nums.len(), which is out of valid index range (max index is len - 1).

2. Step 2: Understand impact on mid + 1 access

   Accessing nums[mid + 1] can cause out-of-bounds panic if mid is at last index due to wrong right boundary.

3. Final Answer:

   Right boundary should be nums.len() - 1 -> Option A

4. Quick Check:

   Right = len - 1 to avoid out-of-bounds [OK]

Hint: Right boundary must be last valid index (len-1) [OK]

Common Mistakes:

• Setting right to length instead of length - 1
• Ignoring out-of-bounds access on mid + 1
• Changing loop condition incorrectly

5. Given a vector nums = vec![1, 2, 3, 1, 5, 6, 4], which index will the binary search peak finding algorithm return and why?

hard

A. 2, because 3 is a peak before the drop

B. 3, because 1 is the middle element

C. 5, because 6 is a peak after the drop

D. 6, because 4 is the last element

Solution

1. Step 1: Identify all peak elements in the array

   Array: [1, 2, 3, 1, 5, 6, 4]. Peaks are elements greater than neighbors: 3 at index 2 and 6 at index 5.

2. Step 2: Trace binary search behavior

   left=0, right=6; mid=3 (1 < 5), left=4; mid=5 (6 > 4), right=5; mid=4 (5 < 6), left=5. Returns index 5.

3. Final Answer:

   5, because 6 is a peak after the drop -> Option C

4. Quick Check:

   Binary search finds peak at index 5 [OK]

Hint: Binary search returns any peak; trace steps to confirm which [OK]

Common Mistakes:

• Assuming it always finds the first peak
• Choosing middle element without checking neighbors
• Ignoring multiple peaks in array