Find Peak Element Using Binary Search in DSA Rust - Build from Scratch

Find Peak Element Using Binary Search

📖 Scenario: Imagine you have a list of numbers representing heights of hills along a trail. You want to find a hill that is taller than its neighbors -- a peak hill. This helps hikers find the highest point nearby quickly.

🎯 Goal: Build a Rust program that finds a peak element in a list of numbers using binary search. A peak element is one that is greater than its neighbors.

📋 What You'll Learn

Create a vector called heights with the exact values: [1, 3, 20, 4, 1, 0]

Create a variable called left and set it to 0

Create a variable called right and set it to the last index of heights

Use a while loop with the condition left < right

Inside the loop, calculate mid as the middle index between left and right

Compare heights[mid] and heights[mid + 1] to decide which side to search next

Print the index of the peak element found

💡 Why This Matters

🌍 Real World

Finding peak elements quickly helps in signal processing, stock price analysis, and terrain mapping where local maximums are important.

💼 Career

Binary search and peak finding algorithms are common in technical interviews and real-world software that requires efficient data analysis.

1

Create the heights vector

Create a vector called heights with the exact values [1, 3, 20, 4, 1, 0]

DSA Rust

# Create the vector heights with values [1, 3, 20, 4, 1, 0]
# Your code here

Hint

Use let heights = vec![1, 3, 20, 4, 1, 0]; to create the vector.

2

Set up left and right pointers

Create variables called left set to 0 and right set to the last index of heights inside the main function

DSA Rust

fn main() {
    let heights = vec![1, 3, 20, 4, 1, 0];
    // Create left and right variables
    // Your code here
}

Hint

Use let mut left = 0; and let mut right = heights.len() - 1;.

3

Implement binary search loop to find peak

Inside the main function, add a while loop with condition left < right. Inside the loop, calculate mid as (left + right) / 2. If heights[mid] is less than heights[mid + 1], set left to mid + 1, else set right to mid

DSA Rust

fn main() {
    let heights = vec![1, 3, 20, 4, 1, 0];
    let mut left = 0;
    let mut right = heights.len() - 1;
    // Add while loop to find peak
    // Your code here
}

Hint

Use a while loop with left < right. Calculate mid and compare heights[mid] with heights[mid + 1].

4

Print the peak element index

After the while loop, print the value of left which is the index of the peak element

DSA Rust

fn main() {
    let heights = vec![1, 3, 20, 4, 1, 0];
    let mut left = 0;
    let mut right = heights.len() - 1;

    while left < right {
        let mid = (left + right) / 2;
        if heights[mid] < heights[mid + 1] {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    // Print the peak index
    // Your code here
}

Hint

Use println!("{}", left); to print the peak index.

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