Binary Search Iterative Approach in DSA Rust - Build from Scratch

Binary Search Iterative Approach

📖 Scenario: You have a sorted list of numbers representing book IDs in a library. You want to find if a specific book ID is available using a fast method.

🎯 Goal: Build a program that uses the iterative binary search method to find a target number in a sorted list.

📋 What You'll Learn

Create a sorted vector of integers called book_ids with the exact values: [10, 20, 30, 40, 50, 60, 70]

Create a variable called target and set it to 40

Write an iterative binary search using a while loop with variables left, right, and mid to find target in book_ids

Print Found at index: X if target is found, or Not found if it is not

💡 Why This Matters

🌍 Real World

Binary search is used in many applications like searching in databases, dictionaries, or any sorted data quickly.

💼 Career

Understanding binary search helps in optimizing search operations and is a common interview question for software developer roles.

1

Create the sorted list of book IDs

Create a vector called book_ids with these exact integers: 10, 20, 30, 40, 50, 60, 70

DSA Rust

# Your code here

Hint

Use let book_ids = vec![10, 20, 30, 40, 50, 60, 70]; to create the vector.

2

Set the target book ID to search

Create a variable called target and set it to 40

DSA Rust

fn main() {
    let book_ids = vec![10, 20, 30, 40, 50, 60, 70];
    // Your code here
}

Hint

Use let target = 40; to set the number you want to find.

3

Write the iterative binary search logic

Write a while loop using variables left, right, and mid to search for target in book_ids iteratively

DSA Rust

fn main() {
    let book_ids = vec![10, 20, 30, 40, 50, 60, 70];
    let target = 40;
    // Your code here
}

Hint

Use while left <= right and calculate mid inside the loop. Compare book_ids[mid] with target and adjust left or right accordingly.

4

Print the search result

Print Found at index: X if target is found at index X, otherwise print Not found

DSA Rust

fn main() {
    let book_ids = vec![10, 20, 30, 40, 50, 60, 70];
    let target = 40;

    let mut left = 0;
    let mut right = book_ids.len() - 1;
    let mut found_index = None;

    while left <= right {
        let mid = left + (right - left) / 2;
        if book_ids[mid] == target {
            found_index = Some(mid);
            break;
        } else if book_ids[mid] < target {
            left = mid + 1;
        } else {
            if mid == 0 { break; } // prevent underflow
            right = mid - 1;
        }
    }

    // Your code here
}

Hint

Use match on found_index to print the correct message.

Practice

1. What is the main purpose of the binary search iterative approach in a sorted array?

easy

A. To find the position of a target value efficiently by repeatedly dividing the search interval in half

B. To sort the array elements in ascending order

C. To reverse the array elements

D. To find the maximum value in the array

Solution

1. Step 1: Understand binary search goal

   Binary search aims to find a target value's position in a sorted array quickly by dividing the search range repeatedly.

2. Step 2: Compare with other options

   Sorting, reversing, or finding max are different operations unrelated to binary search's main goal.

3. Final Answer:

   To find the position of a target value efficiently by repeatedly dividing the search interval in half -> Option A

4. Quick Check:

   Binary search = divide and conquer search [OK]

Hint: Binary search halves the search space each step [OK]

Common Mistakes:

• Confusing binary search with sorting algorithms
• Thinking binary search works on unsorted arrays
• Assuming binary search finds max or min values

2. Which of the following is the correct Rust syntax to start the iterative binary search loop over a sorted slice arr with indices low and high?

easy

A. while low <= high { /* loop body */ }

B. for i in low..high { /* loop body */ }

C. loop { if low > high { break; } /* loop body */ }

D. while low < high { /* loop body */ }

Solution

1. Step 1: Identify correct loop condition for binary search

   Binary search continues while low index is less than or equal to high index to cover all elements.

2. Step 2: Compare options

   while low <= high { /* loop body */ } uses while low <= high, which is correct. for i in low..high { /* loop body */ } uses a for loop which is not suitable. loop { if low > high { break; } /* loop body */ } uses an infinite loop with break, which is less clear. while low < high { /* loop body */ } uses low < high, which misses the case when low equals high.

3. Final Answer:

   while low <= high { /* loop body */ } -> Option A

4. Quick Check:

   Loop condition includes equality for last element [OK]

Hint: Use while low <= high to cover all elements [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using < instead of <= causing missed element check
• Using infinite loop without proper break condition

3. What will be the output of the following Rust code snippet using iterative binary search?

fn binary_search(arr: &[i32], target: i32) -> i32 {
    let mut low = 0;
    let mut high = arr.len() as i32 - 1;
    while low <= high {
        let mid = low + (high - low) / 2;
        if arr[mid as usize] == target {
            return mid;
        } else if arr[mid as usize] < target {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    -1
}

fn main() {
    let arr = [2, 4, 6, 8, 10];
    println!("{}", binary_search(&arr, 6));
}

medium

A. -1

B. 1

C. 2

D. 3

Solution

1. Step 1: Trace binary search for target 6 in array

   Array is [2,4,6,8,10]. Initial low=0, high=4. mid=2. arr[2]=6 equals target, so return 2.

2. Step 2: Confirm returned index

   Index 2 is correct position of 6 in the array.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Check mid index value equals target first [OK]

Common Mistakes:

• Miscomputing mid index causing wrong return
• Returning -1 incorrectly when target exists
• Confusing index with value in output

4. Identify the error in this Rust iterative binary search function and choose the fix:

fn binary_search(arr: &[i32], target: i32) -> i32 {
    let mut low = 0;
    let mut high = arr.len() as i32;
    while low <= high {
        let mid = (low + high) / 2;
        if arr[mid as usize] == target {
            return mid;
        } else if arr[mid as usize] < target {
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }
    -1
}

medium

A. Change while condition to low < high

B. Remove type casting of mid to usize

C. Change mid calculation to mid = low - (high - low) / 2

D. Change high initialization to arr.len() as i32 - 1

Solution

1. Step 1: Analyze high initialization

   High is set to arr.len() which is out of bounds index (since last valid index is length - 1).

2. Step 2: Correct high initialization

   Set high to arr.len() - 1 to avoid index out of bounds error during access.

3. Final Answer:

   Change high initialization to arr.len() as i32 - 1 -> Option D

4. Quick Check:

   High must be last valid index [OK]

Hint: High index = length - 1 to avoid out of bounds [OK]

Common Mistakes:

• Setting high to length causing panic
• Changing while condition incorrectly
• Incorrect mid calculation formula

5. Given a sorted array arr = [1, 3, 5, 7, 9, 11], write an iterative binary search function in Rust to find the index of the smallest element greater than or equal to a target value. What will be the output of your function when searching for target 8?

hard

A. 5

B. 4

C. -1

D. 3

Solution

1. Step 1: Understand problem - find smallest element >= target

   We want the index of the first element not less than target 8.

2. Step 2: Trace binary search for target 8

   Array: [1,3,5,7,9,11]. Elements >=8 are 9 (index 4) and 11 (index 5). The smallest is 9 at index 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   First element >= 8 is 9 at index 4 [OK]

Hint: Find first element >= target by adjusting high on match [OK]

Common Mistakes:

• Returning index of element just less than target
• Returning -1 when element exists
• Confusing smallest element greater than target with equal