Binary Search as Divide and Conquer in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Search on Sorted Array

What is the output of this Go code that performs binary search to find the index of number 7 in the sorted array?

DSA Go

package main
import "fmt"

func binarySearch(arr []int, target int) int {
    left, right := 0, len(arr)-1
    for left <= right {
        mid := left + (right-left)/2
        if arr[mid] == target {
            return mid
        } else if arr[mid] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return -1
}

func main() {
    arr := []int{1, 3, 5, 7, 9, 11}
    fmt.Println(binarySearch(arr, 7))
}

A 3

B 2

C -1

D 4

💡 Hint

Remember that array indices start at 0 and 7 is the fourth element.

The array is [1,3,5,7,9,11]. The number 7 is at index 3 (0-based). The binary search returns this index.

❓ Predict Output

intermediate

Binary Search Result When Target Not Present

What will be the output of this Go code when searching for number 8 in the sorted array?

DSA Go

package main
import "fmt"

func binarySearch(arr []int, target int) int {
    left, right := 0, len(arr)-1
    for left <= right {
        mid := left + (right-left)/2
        if arr[mid] == target {
            return mid
        } else if arr[mid] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return -1
}

func main() {
    arr := []int{2, 4, 6, 10, 12}
    fmt.Println(binarySearch(arr, 8))
}

A 3

B 2

C -1

D 0

💡 Hint

If the target is not found, binary search returns -1.

The number 8 is not in the array, so the function returns -1.

🧠 Conceptual

advanced

Divide and Conquer Principle in Binary Search

Which statement best describes how binary search uses the divide and conquer approach?

A It divides the array into two halves and recursively searches only the half that may contain the target.

B It searches all elements in the array simultaneously using parallel processing.

C It divides the array into multiple parts and searches each part independently before combining results.

D It sorts the array first and then performs a linear search on the entire array.

💡 Hint

Think about how binary search reduces the search space by half each step.

Binary search splits the array into two halves and continues searching only in the half where the target can be, ignoring the other half.

🔧 Debug

advanced

Identify the Bug in Binary Search Implementation

What error will this Go code produce when running binary search, and why?

DSA Go

package main
import "fmt"

func binarySearch(arr []int, target int) int {
    left, right := 0, len(arr)-1
    for left <= right {
        mid := left + (right-left)/2
        if arr[mid] == target {
            return mid
        } else if arr[mid] < target {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return -1
}

func main() {
    arr := []int{1, 2, 3, 4, 5}
    fmt.Println(binarySearch(arr, 6))
}

A compiles but runs infinitely

B runtime error: index out of range

C returns -1 even if target is present

D prints -1 correctly

💡 Hint

Check the initial value of 'right' and how it relates to array indices.

The variable 'right' is set to len(arr), which is out of bounds for indexing. Accessing arr[mid] when mid equals len(arr) causes a runtime error.

🚀 Application

expert

Binary Search on Rotated Sorted Array

Given a rotated sorted array, which option correctly finds the index of target 3 using a modified binary search?

DSA Go

package main
import "fmt"

func searchRotated(arr []int, target int) int {
    left, right := 0, len(arr)-1
    for left <= right {
        mid := left + (right-left)/2
        if arr[mid] == target {
            return mid
        }
        if arr[left] <= arr[mid] {
            if arr[left] <= target && target < arr[mid] {
                right = mid - 1
            } else {
                left = mid + 1
            }
        } else {
            if arr[mid] < target && target <= arr[right] {
                left = mid + 1
            } else {
                right = mid - 1
            }
        }
    }
    return -1
}

func main() {
    arr := []int{4, 5, 6, 7, 0, 1, 2, 3}
    fmt.Println(searchRotated(arr, 3))
}

A 0

B 3

C -1

D 7

💡 Hint

Remember the array is rotated; index 7 holds the number 3.

The target 3 is at index 7 in the rotated array. The modified binary search correctly finds and returns this index.

Practice

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

easy

A. It uses recursion to add elements to the array.

B. It searches every element one by one until the target is found.

C. It divides the search range in half each time to find the target efficiently.

D. It sorts the array before searching for the target.

Solution

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

   Binary search splits the search space into two halves and discards one half each time.

2. Step 2: Identify the efficiency of halving the search space

   By halving the range repeatedly, it quickly narrows down to the target element.

3. Final Answer:

   It divides the search range in half each time to find the target efficiently. -> Option C

4. Quick Check:

   Divide and conquer means halving search space [OK]

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

Common Mistakes:

• Thinking binary search checks every element
• Confusing sorting with searching
• Believing binary search adds elements

2. Which of the following is the correct Go syntax to start a binary search loop on a sorted slice arr?

easy

A. for left := 0; left < len(arr); left++ { }

B. for i := 0; i < len(arr); i++ { }

C. while left < right { }

D. for left, right := 0, len(arr)-1; left <= right; { }

Solution

1. Step 1: Identify Go for-loop syntax for binary search

   Binary search uses two pointers, left and right, initialized to start and end indices.

2. Step 2: Confirm loop condition and initialization

   The loop continues while left <= right, using Go's for syntax with multiple variables.

3. Final Answer:

   for left, right := 0, len(arr)-1; left <= right; { } -> Option D

4. Quick Check:

   Go for-loop with two vars and condition [OK]

Hint: Binary search loop uses two pointers and condition left <= right [OK]

Common Mistakes:

• Using for-loop with single index only
• Using while syntax (not valid in Go)
• Incorrect loop condition like left < right

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

arr := []int{2, 4, 6, 8, 10}
target := 6
left, right := 0, len(arr)-1
found := false
for left <= right {
  mid := left + (right-left)/2
  if arr[mid] == target {
    found = true
    break
  } else if arr[mid] < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}
fmt.Println(found)

medium

A. false

B. true

C. 6

D. runtime error

Solution

1. Step 1: Trace the binary search steps for target 6

   Initial left=0, right=4, mid=2 (arr[2]=6), matches target, so found=true.

2. Step 2: Confirm loop breaks and prints found

   Loop breaks immediately after finding target, so prints true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Target found at mid index 2 [OK]

Hint: Check mid element equals target first [OK]

Common Mistakes:

• Not updating left or right correctly
• Off-by-one errors in mid calculation
• Assuming output is the target value

4. Identify the bug in this binary search Go code snippet:

arr := []int{1, 3, 5, 7, 9}
target := 4
left, right := 0, len(arr)
for left <= right {
  mid := (left + right) / 2
  if arr[mid] == target {
    fmt.Println("Found")
    break
  } else if arr[mid] < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}

medium

A. right should be initialized to len(arr)-1, not len(arr)

B. mid calculation should be mid := left - right

C. left should be incremented by 2 instead of 1

D. The loop condition should be left < right

Solution

1. Step 1: Check initialization of right pointer

   right is set to len(arr), which is out of valid index range (max index is len(arr)-1).

2. Step 2: Correct right initialization and explain impact

   Setting right to len(arr)-1 fixes out-of-range error when accessing arr[mid].

3. Final Answer:

   right should be initialized to len(arr)-1, not len(arr) -> Option A

4. Quick Check:

   Array indices go from 0 to len(arr)-1 [OK]

Hint: Initialize right to last valid index len(arr)-1 [OK]

Common Mistakes:

• Using len(arr) as index causes runtime error
• Wrong mid calculation formula
• Incorrect loop condition causing infinite loop

5. Given a sorted slice arr := []int{1, 2, 2, 2, 3, 4}, how can you modify binary search to find the first occurrence of 2?

hard

A. After finding 2, continue searching left half to find earlier occurrence

B. Return immediately when 2 is found at any position

C. Search the right half after finding 2 to find last occurrence

D. Use linear search instead of binary search

Solution

1. Step 1: Understand the problem of duplicates in binary search

   Standard binary search returns any occurrence, not necessarily the first.

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

   When target found, update result and move right pointer to mid-1 to find earlier occurrence.

3. Final Answer:

   After finding 2, continue searching left half to find earlier occurrence -> Option A

4. Quick Check:

   Find first occurrence by searching left side after match [OK]

Hint: Keep searching left half after finding target to get first occurrence [OK]

Common Mistakes:

• Stopping at first found occurrence
• Searching right half instead of left
• Using linear search unnecessarily