Binary Search Iterative Approach in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Binary Search for Existing Element

What is the output of the following Go code that performs an iterative binary search for the number 7 in a sorted array?

DSA Go

package main
import "fmt"

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

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

A -1

B 2

C 3

D 4

💡 Hint

Remember that array indices start at 0 and the array is sorted.

The number 7 is at index 3 in the array [1, 3, 5, 7, 9, 11]. The binary search returns the index where the target is found.

❓ Predict Output

intermediate

Output of Binary Search for Non-Existing Element

What is the output of the following Go code that performs an iterative binary search for the number 7 in a sorted array?

DSA Go

package main
import "fmt"

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

func main() {
    arr := []int{2, 4, 6, 8, 10, 12}
    fmt.Println(binarySearch(arr, 7))
}

A 4

B 3

C 2

D -1

💡 Hint

Check if the target exists in the array and note its index.

The number 7 is not present in the array [2, 4, 6, 8, 10, 12]. The binary search returns -1 when the target is not found.

❓ Predict Output

advanced

Output of Binary Search with Duplicate Elements

What is the output of the following Go code that performs an iterative binary search for the number 5 in a sorted array with duplicates?

DSA Go

package main
import "fmt"

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

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

A 2

B 3

C 4

D -1

💡 Hint

Binary search may return any index where the target is found.

The binary search returns index 3 where 5 is found (among indices 2,3,4). It may not always return the first or last duplicate but a valid index.

🧠 Conceptual

advanced

Why Use Iterative Binary Search Instead of Recursive?

Which of the following is the main advantage of using an iterative approach for binary search over a recursive one?

A Iterative approach uses less memory by avoiding function call stack overhead.

B Iterative approach is easier to write and understand than recursive.

C Iterative approach can handle unsorted arrays.

D Iterative approach always finds the first occurrence of duplicates.

💡 Hint

Think about memory usage and function calls.

Iterative binary search avoids the overhead of recursive calls, saving memory and reducing risk of stack overflow.

🔧 Debug

expert

Identify the Bug in Binary Search Implementation

What error will the following Go code produce when running binary search on the array [1, 2, 3, 4, 5] searching for 6?

DSA Go

package main
import "fmt"

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

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

A runtime error: index out of range

B returns 2 (correct index)

C returns -1 (not found)

D compilation error

💡 Hint

Check the initial value of 'high' and array indexing.

The variable 'high' is set to len(arr), which is out of valid index range (max index is len(arr)-1). Accessing arr[mid] can cause index out of range error.

Practice

1. What is the main purpose of the mid variable in the iterative binary search algorithm?

easy

A. To keep track of the number of iterations

B. To store the target value to search

C. To find the middle index of the current search range

D. To hold the entire array being searched

Solution

1. Step 1: Understand the role and importance of mid in binary search

   The mid variable calculates the middle index between the current low and high indices to split the search range. By checking the element at mid, the algorithm decides whether to search the left or right half next, making the search efficient.

2. Final Answer:

   To find the middle index of the current search range -> Option C

3. Quick Check:

   mid = middle index [OK]

Hint: Mid splits search range in half each step [OK]

Common Mistakes:

• Confusing mid with target value
• Using mid as a counter instead of index
• Not updating mid inside the loop

2. Which of the following is the correct way to calculate the middle index mid in Go to avoid integer overflow?

easy

A. mid := (low + high) / 2

B. mid := (high - low) / 2

C. mid := (low - high) / 2

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

Solution

1. Step 1: Understand integer overflow risk and safe calculation

   Adding low + high directly can overflow if both are large integers. Calculating mid as low + (high - low) / 2 avoids overflow by subtracting first.

2. Final Answer:

   mid := low + (high - low) / 2 -> Option D

3. Quick Check:

   Safe mid calculation avoids overflow [OK]

Hint: Use low + (high - low)/2 to prevent overflow [OK]

Common Mistakes:

• Using (low + high)/2 directly causing overflow
• Subtracting incorrectly causing negative mid
• Confusing order of operations

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

package main
import "fmt"

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

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

medium

A. 3

B. 2

C. -1

D. 0

Solution

1. Step 1: Trace the binary search for target 6 and verify return

   Array is [2,4,6,8,10]. Initial low=0, high=4. mid=2. arr[2]=6 matches target. Function returns mid = 2 immediately.

2. Final Answer:

   2 -> Option B

3. Quick Check:

   Index of 6 in array = 2 [OK]

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

Common Mistakes:

• Returning value instead of index
• Off-by-one errors in mid calculation
• Not updating low/high correctly

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

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

medium

A. The initial value of high should be len(arr) - 1

B. The loop condition should be low < high

C. The mid calculation causes overflow

D. The function should return arr[mid] instead of mid

Solution

1. Step 1: Check high initialization and correct it

   High is set to len(arr), which is out of array index range (max index is len(arr)-1). Set high to len(arr) - 1 to avoid index out of range error when accessing arr[mid].

2. Final Answer:

   The initial value of high should be len(arr) - 1 -> Option A

3. Quick Check:

   High = len(arr)-1 fixes index error [OK]

Hint: High index must be last valid array index [OK]

Common Mistakes:

• Setting high to len(arr) causing runtime error
• Using wrong loop condition
• Returning value instead of index

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

hard

A. When arr[mid] == target, update high = mid - 1 and continue searching

B. Return mid immediately when arr[mid] == target

C. When arr[mid] == target, update low = mid + 1 and continue searching

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

Solution

1. Step 1: Understand duplicates issue and modify search logic

   Standard binary search returns any occurrence, not necessarily the first. When arr[mid] == target, move high to mid - 1 to continue searching left side for earlier occurrence, but store mid as potential answer.

2. Final Answer:

   When arr[mid] == target, update high = mid - 1 and continue searching -> Option A

3. Quick Check:

   Adjust high to find first occurrence [OK]

Hint: Move high left on match to find first occurrence [OK]

Common Mistakes:

• Returning immediately on first match
• Moving low right on match losing first occurrence
• Using linear search unnecessarily