Binary Search Recursive Approach in DSA Go

Mental Model

Binary search splits a sorted list in half to find a target quickly by checking the middle element and searching only the half that could contain the target.

Analogy: Imagine looking for a word in a dictionary by opening it in the middle, then deciding if you should look in the first half or the second half, repeating this until you find the word.

Array: [1, 3, 5, 7, 9, 11, 13]
Indexes:  0   1   2   3   4    5    6
          ↑               ↑
       low=0           high=6
          ↑
       mid=3 (value=7)

Dry Run Walkthrough

Input: list: [1, 3, 5, 7, 9, 11, 13], target: 9

Goal: Find the index of target 9 using recursive binary search

Step 1: Calculate mid index between low=0 and high=6; mid=3, value=7

[1, 3, 5, [7], 9, 11, 13] low=0 high=6 mid=3

Why: Check middle value to decide which half to search next

Step 2: Target 9 is greater than mid value 7, search right half: low=4, high=6

[1, 3, 5, 7, [9], 11, 13] low=4 high=6

Why: Target must be in the right half because 9 > 7

Step 3: Calculate mid index between low=4 and high=6; mid=5, value=11

[1, 3, 5, 7, 9, [11], 13] low=4 high=6 mid=5

Why: Check middle value of the new subarray

Step 4: Target 9 is less than mid value 11, search left half: low=4, high=4

[1, 3, 5, 7, [9], 11, 13] low=4 high=4

Why: Target must be in the left half because 9 < 11

Step 5: Calculate mid index between low=4 and high=4; mid=4, value=9

[1, 3, 5, 7, [9], 11, 13] low=4 high=4 mid=4

Why: Check the only element left

Step 6: Found target 9 at index 4, return index

[1, 3, 5, 7, [9], 11, 13]

Why: Target matches mid value, search ends

Result:

Final state: [1, 3, 5, 7, [9], 11, 13] Target found at index 4

Annotated Code

DSA Go

package main

import "fmt"

func binarySearchRecursive(arr []int, target int, low int, high int) int {
    if low > high {
        return -1 // target not found
    }
    mid := low + (high-low)/2
    if arr[mid] == target {
        return mid // target found
    } else if arr[mid] > target {
        return binarySearchRecursive(arr, target, low, mid-1) // search left half
    } else {
        return binarySearchRecursive(arr, target, mid+1, high) // search right half
    }
}

func main() {
    arr := []int{1, 3, 5, 7, 9, 11, 13}
    target := 9
    index := binarySearchRecursive(arr, target, 0, len(arr)-1)
    if index != -1 {
        fmt.Printf("Target %d found at index %d\n", target, index)
    } else {
        fmt.Printf("Target %d not found\n", target)
    }
}

if low > high {

stop recursion if search range is invalid (target not found)

mid := low + (high-low)/2

calculate middle index to split search range

if arr[mid] == target {

check if middle element is the target

return binarySearchRecursive(arr, target, low, mid-1)

search left half if target is smaller than mid value

return binarySearchRecursive(arr, target, mid+1, high)

search right half if target is greater than mid value

Output

Target 9 found at index 4

Complexity Analysis

Time: O(log n) because each recursive call halves the search range

Space: O(log n) due to recursive call stack depth proportional to log n

vs Alternative: Compared to linear search O(n), binary search is much faster on sorted arrays by reducing search space quickly

Edge Cases

empty array

function returns -1 immediately as low > high

DSA Go

if low > high {

target not in array

search narrows until low > high, then returns -1

DSA Go

if low > high {

single element array with target present

mid equals low equals high, target found or not found accordingly

DSA Go

if arr[mid] == target {

When to Use This Pattern

When you need to find an item quickly in a sorted list, use recursive binary search because it efficiently halves the search space each step.

Common Mistakes

Mistake: Calculating mid as (low + high) can cause integer overflow
Fix: Calculate mid as low + (high - low) / 2 to avoid overflow

Mistake: Not updating low or high correctly in recursive calls
Fix: Ensure recursive calls use mid-1 or mid+1 to shrink search range properly

Summary

It finds a target value in a sorted array by recursively checking the middle element and searching the half that may contain the target.

Use it when you have a sorted list and want to find an element quickly.

The key insight is that each step cuts the search space in half, making the search very fast.

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