Binary Search Iterative Approach in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the binary search iterative approach?

Binary search iterative approach repeatedly divides the sorted list in half to find the target value without using recursion.

beginner

In binary search iterative approach, what happens if the middle element is less than the target?

If the middle element is less than the target, the search continues in the right half of the list by moving the start pointer to middle + 1.

beginner

Why is the binary search iterative approach efficient compared to linear search?

Because it reduces the search space by half each time, making the time complexity O(log n), which is much faster than linear search's O(n).

beginner

What are the key variables used in the binary search iterative approach?

The key variables are start (beginning index), end (ending index), and mid (middle index) to track the current search range.

intermediate

Show the Rust code snippet for the binary search iterative approach.

fn binary_search(arr: &[i32], target: i32) -> Option<usize> {
    if arr.is_empty() {
        return None;
    }
    let mut start = 0;
    let mut end = arr.len() - 1;
    while start <= end {
        let mid = start + (end - start) / 2;
        if arr[mid] == target {
            return Some(mid);
        } else if arr[mid] < target {
            start = mid + 1;
        } else {
            if mid == 0 { break; } // prevent underflow
            end = mid - 1;
        }
    }
    None
}

What is the time complexity of the binary search iterative approach?

A O(log n)

B O(n)

C O(n log n)

D O(1)

Binary search divides the search space in half each step, resulting in O(log n) time complexity.

In binary search iterative approach, if the target is less than the middle element, what should you do?

A Search the right half

B Return the middle index

C Search the left half

D Stop searching

If target is less than middle element, search continues in the left half by moving the end pointer.

Which variable represents the middle index in binary search iterative approach?

A start

B end

C target

D mid

The 'mid' variable holds the middle index between start and end.

What condition ends the binary search iterative loop?

A start == end

B start > end

C mid == target

D target not found

The loop ends when start pointer passes end pointer, meaning the target is not found.

What does the binary search iterative function return if the target is not found?

A None

B Some(index)

C 0

D -1

The function returns None to indicate the target is not present in the array.

Explain how the binary search iterative approach works step-by-step.

Think about how you cut the search space in half each time.

Write the Rust code for binary search iterative approach and explain each part.

Focus on the loop and how pointers move.

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