Binary Search as Divide and Conquer in DSA Rust - Time & Space Complexity

Time Complexity: Binary Search as Divide and Conquer

O(log n)

Understanding Time Complexity

We want to understand how fast binary search finds a number in a sorted list.

How does the time it takes change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code looks for a target number by repeatedly cutting the search area in half.

Identify Repeating Operations

• Primary operation: Checking the middle element of the current search range.
• How many times: The search range halves each time until the target is found or range is empty.

How Execution Grows With Input

Each step cuts the list size in half, so the number of steps grows slowly as the list grows.

Input Size (n)	Approx. Operations
10	About 4 checks
100	About 7 checks
1000	About 10 checks

Pattern observation: Doubling the list size adds only one extra check.

Final Time Complexity

Time Complexity: O(log n)

This means the time to find the target grows slowly, adding just a few more steps even if the list gets much bigger.

Common Mistake

[X] Wrong: "Binary search checks every element one by one like linear search."

[OK] Correct: Binary search skips half the list each time, so it does far fewer checks than looking at every element.

Interview Connect

Knowing how binary search quickly narrows down answers shows you understand efficient searching, a key skill in many coding problems.

Self-Check

"What if the list was not sorted? How would the time complexity of searching change?"

Practice

1. What is the main idea behind Binary Search as a divide and conquer algorithm?

easy

A. Sort the list first, then search linearly

B. Check every element one by one until the target is found

C. Repeatedly split the sorted list in half to find the target

D. Use a hash table to find the target instantly

Solution

1. Step 1: Understand divide and conquer in binary search

   Binary search works by dividing the sorted list into two halves and deciding which half to search next.

2. Step 2: Identify the correct approach

   It repeatedly splits the list in half, narrowing down the search space until the target is found or no elements remain.

3. Final Answer:

   Repeatedly split the sorted list in half to find the target -> Option C

4. Quick Check:

   Divide and conquer = split list halves [OK]

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

Common Mistakes:

• Thinking binary search checks every element
• Assuming list must be unsorted
• Confusing binary search with linear search

2. Which of the following is the correct Rust function signature for a binary search function that returns an Option<usize>?

easy

A. fn binary_search(arr: &[i32], target: i32) -> Option<usize>

B. fn binary_search(arr: Vec bool

C. fn binary_search(arr: &[i32], target: i32) -> usize

D. fn binary_search(arr: &mut [i32], target: i32) -> i32

Solution

1. Step 1: Understand Rust function signature for binary search

   The function takes a slice of integers and a target integer, returning an Option with the index if found.

2. Step 2: Match the correct signature

   Option<usize> is used to return Some(index) if found or None if not found, matching fn binary_search(arr: &[i32], target: i32) -> Option<usize>.

3. Final Answer:

   fn binary_search(arr: &[i32], target: i32) -> Option<usize> -> Option A

4. Quick Check:

   Return Option<usize> for index or None [OK]

Hint: Use Option<usize> to handle found or not found [OK]

Common Mistakes:

• Returning usize without Option causes errors if not found
• Using Vec instead of slice for input unnecessarily
• Returning bool loses index information

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

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

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

medium

A. Some(3)

B. Some(2)

C. None

D. Some(1)

Solution

1. Step 1: Trace binary search on array [2,4,6,8,10] for target 6

   Initial left=0, right=4, mid=2. arr[2]=6 matches target.

2. Step 2: Return index 2 wrapped in Some

   Since arr[mid] == target, function returns Some(2).

3. Final Answer:

   Some(2) -> Option B

4. Quick Check:

   Index of 6 in array is 2 [OK]

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

Common Mistakes:

• Off-by-one errors in mid calculation
• Confusing index with value
• Returning None incorrectly when target exists

4. Identify the bug in this binary search Rust code and choose the correct fix:

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

medium

A. Change while condition to left < right

B. Change mid calculation to (left + right) / 2 without subtraction

C. Remove else if and use only if and else

D. Change right initialization to arr.len() - 1

Solution

1. Step 1: Analyze right boundary initialization

   Right is set to arr.len(), which is out of bounds since indexing is 0-based.

2. Step 2: Fix right to arr.len() - 1 to avoid out-of-bounds access

   Setting right to arr.len() - 1 ensures mid index stays within valid range.

3. Final Answer:

   Change right initialization to arr.len() - 1 -> Option D

4. Quick Check:

   Right boundary must be last valid index [OK]

Hint: Right index = length - 1 for zero-based arrays [OK]

Common Mistakes:

• Using arr.len() as right causes panic
• Changing while condition incorrectly
• Incorrect mid calculation causing overflow

5. Given a sorted array with possible duplicate values, how can you modify binary search to find the first occurrence of a target value?

hard

A. After finding target at mid, continue searching left half to find earlier occurrence

B. Return immediately when target is found at mid

C. Search right half only after finding target

D. Use linear search after binary search to find first occurrence

Solution

1. Step 1: Understand duplicates and first occurrence requirement

   Binary search normally returns any matching index, but to find first occurrence, we must check if earlier duplicates exist.

2. Step 2: Modify search to continue left after finding target

   When target is found at mid, record index and continue searching left half to find if an earlier occurrence exists.

3. Final Answer:

   After finding target at mid, continue searching left half to find earlier occurrence -> Option A

4. Quick Check:

   Find first occurrence by searching left after match [OK]

Hint: Keep searching left half after finding target [OK]

Common Mistakes:

• Stopping search immediately on first match
• Searching right half instead of left
• Using linear search unnecessarily