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

Time Complexity: Binary Search Recursive Approach

O(log n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func binarySearch(arr []int, target, low, high int) int {
    if low > high {
        return -1
    }
    mid := low + (high - low) / 2
    if arr[mid] == target {
        return mid
    } else if arr[mid] > target {
        return binarySearch(arr, target, low, mid-1)
    } else {
        return binarySearch(arr, target, mid+1, high)
    }
}

This code searches for a target number in a sorted list 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: The search range is halved each call, repeating until the range is empty or target found.

How Execution Grows With Input

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

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

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

Final Time Complexity

Time Complexity: O(log n)

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

Common Mistake

[X] Wrong: "Binary search checks every item one by one, so it takes as long as the list size."

[OK] Correct: Binary search skips half the list each time, so it does not check every item, making it much faster.

Interview Connect

Understanding binary search time helps you explain efficient searching clearly and confidently in interviews.

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 the binary search recursive approach on a sorted array?

easy

A. Sort the array first and then search linearly

B. Divide the array into halves and search recursively in the half where the target may exist

C. Search each element one by one from start to end

D. Use a queue to store elements and search iteratively

Solution

1. Step 1: Understand binary search logic

   Binary search works by dividing the sorted array into two halves and deciding which half to search next based on comparison with the middle element.

2. Step 2: Apply recursion in binary search

   The recursive approach calls the same function on the half where the target might be, reducing the search space each time.

3. Final Answer:

   Divide the array into halves and search recursively in the half where the target may exist -> Option B

4. Quick Check:

   Binary search = divide and conquer [OK]

Hint: Remember: binary search splits array and searches half recursively [OK]

Common Mistakes:

• Trying to search linearly instead of dividing
• Not using recursion properly
• Ignoring that array must be sorted

2. Which of the following is the correct function signature for a recursive binary search in Go?

easy

A. func binarySearch(arr []int, target int, mid int) int

B. func binarySearch(arr []int, target int) int

C. func binarySearch(arr []int, target int, low int, high int) int

D. func binarySearch(arr []int) int

Solution

1. Step 1: Identify parameters needed for recursion

   Recursive binary search needs the array, target, and the current low and high indices to know the search range.

2. Step 2: Check function signatures

   func binarySearch(arr []int, target int, low int, high int) int includes low and high indices, which are essential for recursion. Others miss these or have incorrect parameters.

3. Final Answer:

   func binarySearch(arr []int, target int, low int, high int) int -> Option C

4. Quick Check:

   Parameters include low and high indices [OK]

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

Common Mistakes:

• Omitting low and high indices in parameters
• Adding unnecessary parameters like mid
• Using wrong return types

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

func binarySearch(arr []int, target, low, high int) int {
    if low > high {
        return -1
    }
    mid := (low + high) / 2
    if arr[mid] == target {
        return mid
    } else if arr[mid] > target {
        return binarySearch(arr, target, low, mid-1)
    } else {
        return binarySearch(arr, target, mid+1, high)
    }
}

func main() {
    arr := []int{2, 4, 6, 8, 10}
    result := binarySearch(arr, 6, 0, len(arr)-1)
    fmt.Println(result)
}

medium

A. 0

B. 3

C. -1

D. 2

Solution

1. Step 1: Trace the recursive calls for target 6

   Initial call: low=0, high=4, mid=2, arr[2]=6 equals target, so return 2 immediately.

2. Step 2: Confirm output

   Since target found at index 2, output is 2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Target 6 found at index 2 [OK]

Hint: Check mid index value first to find target quickly [OK]

Common Mistakes:

• Confusing index with value
• Not updating low and high correctly
• Returning wrong index or -1 incorrectly

4. Identify the error in the following recursive binary search function and select the fix:

func binarySearch(arr []int, target, low, high int) int {
    if low >= high {
        return -1
    }
    mid := (low + high) / 2
    if arr[mid] == target {
        return mid
    } else if arr[mid] > target {
        return binarySearch(arr, target, low, mid-1)
    } else {
        return binarySearch(arr, target, mid+1, high)
    }
}

medium

A. Change condition to if low > high instead of low >= high

B. Change mid calculation to mid := (low + high + 1) / 2

C. Return 0 instead of -1 when not found

D. Remove else if and use only if statements

Solution

1. Step 1: Analyze base case condition

   The base case should stop recursion when low is greater than high, meaning no elements left to search. Using low >= high skips searching when low equals high, missing that element.

2. Step 2: Correct the condition

   Changing to low > high allows searching the last element when low equals high.

3. Final Answer:

   Change condition to if low > high instead of low >= high -> Option A

4. Quick Check:

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

Hint: Base case should be low > high, not low >= high [OK]

Common Mistakes:

• Stopping recursion too early with wrong base case
• Incorrect mid calculation causing infinite recursion
• Returning wrong default value when not found

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 found, check if previous element is same; if yes, search left half recursively

B. Return the first found index immediately without further checks

C. Search only the right half recursively after finding target

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

Solution

1. Step 1: Understand duplicates in binary search

   Standard binary search returns any occurrence of target. To find first occurrence, we must check if earlier elements are also equal.

2. Step 2: Modify recursion to check left side

   When target found at mid, check if mid is first index or previous element is different. If previous element is same, recursively search left half to find earlier occurrence.

3. Final Answer:

   When target found, check if previous element is same; if yes, search left half recursively -> Option A

4. Quick Check:

   Check left side for first occurrence [OK]

Hint: On finding target, move left if previous equals target [OK]

Common Mistakes:

• Returning first found index without checking duplicates
• Searching right half instead of left for first occurrence
• Using linear search unnecessarily