Binary Search Recursive Approach in DSA Rust - Time & Space Complexity

Time Complexity: Binary Search Recursive Approach

O(log n)

Understanding Time Complexity

We want to understand how the time taken by recursive binary search changes as the input size grows.

Specifically, how many steps does it take to find a value in a sorted list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> {
    if low > high {
        return None;
    }
    let mid = low + (high - low) / 2;
    if arr[mid] == target {
        Some(mid)
    } else if arr[mid] < target {
        binary_search(arr, target, mid + 1, high)
    } else {
        binary_search(arr, target, low, mid.saturating_sub(1))
    }
}

This code searches for a target number in a sorted array by repeatedly dividing the search range in half.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call that halves the search range each time.
• How many times: At most, the function calls itself until the range is empty, roughly halving the size each time.

How Execution Grows With Input

Each step cuts the search area in half, so the number of steps grows slowly as the list gets bigger.

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

Pattern observation: Doubling the input size adds only one extra step.

Final Time Complexity

Time Complexity: O(log n)

This means the number of steps grows slowly, increasing by one each time the input size doubles.

Common Mistake

[X] Wrong: "Binary search checks every element one by one, so it is O(n)."

[OK] Correct: Binary search does not check every element; it splits the search area in half each time, so it only needs a few steps even for large lists.

Interview Connect

Understanding binary search's time complexity shows you can analyze efficient algorithms and explain why they work well, a key skill in interviews.

Self-Check

"What if we changed the array to be unsorted? How would the time complexity change?"

Practice

1. What is the main idea behind the binary search recursive approach?

easy

A. Divide the search range in half recursively until the target is found or range is empty

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

C. Sort the array recursively before searching

D. Use a queue to store elements and search iteratively

Solution

1. Step 1: Understand binary search logic

   Binary search works by repeatedly dividing the search range in half to find the target efficiently.

2. Step 2: Recognize recursive approach

   The recursive approach calls the same function on the half where the target might be, until found or range is empty.

3. Final Answer:

   Divide the search range in half recursively until the target is found or range is empty -> Option A

4. Quick Check:

   Binary search recursive = divide and conquer [OK]

Hint: Binary search halves range recursively to find target fast [OK]

Common Mistakes:

• Thinking binary search checks every element
• Confusing recursion with iteration
• Assuming array must be sorted during search

2. Which of the following is the correct Rust function signature for a recursive binary search on a slice of integers?

easy

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

B. fn binary_search(arr: Vec usize

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

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

Solution

1. Step 1: Identify parameters for recursion

   Recursive binary search needs the array slice, target, and current low and high indices to narrow search.

2. Step 2: Match correct signature

   fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> matches this with slice reference, target, low and high indices, and returns Option with found index.

3. Final Answer:

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

4. Quick Check:

   Recursive binary search needs low and high indices [OK]

Hint: Recursive binary search needs low and high indices parameters [OK]

Common Mistakes:

• Using Vec instead of slice reference
• Missing low and high parameters for recursion
• Returning wrong type like bool or i32

3. What will be the output of the following Rust code snippet?

fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> {
    if low > high {
        return None;
    }
    let mid = low + (high - low) / 2;
    if arr[mid] == target {
        return Some(mid);
    } else if arr[mid] > target {
        return binary_search(arr, target, low, mid - 1);
    } else {
        return binary_search(arr, target, mid + 1, high);
    }
}

fn main() {
    let arr = [2, 4, 6, 8, 10];
    let result = binary_search(&arr, 6, 0, arr.len() - 1);
    println!("{:?}", result);
}

medium

A. Some(3)

B. Some(2)

C. None

D. Some(1)

Solution

1. Step 1: Trace initial call and mid calculation

   Initial call: low=0, high=4, mid=2. arr[2] = 6 matches target 6.

2. Step 2: Return index wrapped in Some

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

3. Final Answer:

   Some(2) -> Option B

4. Quick Check:

   Target found at index 2 [OK]

Hint: Calculate mid index and compare target to find result fast [OK]

Common Mistakes:

• Miscomputing mid index
• Confusing index with value
• Forgetting zero-based indexing

4. Identify the error in this recursive binary search Rust function:

fn binary_search(arr: &[i32], target: i32, low: usize, high: usize) -> Option<usize> {
    if low >= high {
        return None;
    }
    let mid = low + (high - low) / 2;
    if arr[mid] == target {
        return Some(mid);
    } else if arr[mid] > target {
        return binary_search(arr, target, low, mid - 1);
    } else {
        return binary_search(arr, target, mid + 1, high);
    }
}

medium

A. The recursive calls should swap low and high parameters

B. Mid calculation should use low + (high - low) / 2 to avoid overflow

C. The function should return usize instead of Option<usize>

D. The base case condition should be low > high, not low >= high

Solution

1. Step 1: Analyze base case condition

   Base case currently returns None if low >= high, but when low == high, the element at that index should still be checked.

2. Step 2: Correct base case to low > high

   Changing condition to low > high ensures the last element is checked before returning None.

3. Final Answer:

   The base case condition should be low > high, not low >= high -> Option D

4. Quick Check:

   Base case must allow low == high check [OK]

Hint: Base case must allow checking when low equals high [OK]

Common Mistakes:

• Stopping search too early by wrong base case
• Ignoring possibility of single element left
• Misusing mid calculation causing overflow

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

hard

A. When target is found, continue searching left half recursively to find earlier occurrence

B. Return immediately when target is found without further search

C. Use iterative binary search instead of recursive

D. Sort the array again before searching

Solution

1. Step 1: Understand duplicates and first occurrence

   Duplicates mean target may appear multiple times; first occurrence is the smallest index with target.

2. Step 2: Modify recursion to search left half after finding target

   When target is found at mid, recursively search left half to check for earlier occurrence, else return mid.

3. Final Answer:

   When target is found, continue searching left half recursively to find earlier occurrence -> Option A

4. Quick Check:

   Find first occurrence by searching left after match [OK]

Hint: After finding target, search left half to find first occurrence [OK]

Common Mistakes:

• Stopping at first found index ignoring duplicates
• Sorting array again unnecessarily
• Switching to iterative without reason