Binary Search Iterative Approach in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the middle index in the binary search loop.

DSA Go

mid := low [1] (high - low) / 2

Drag options to blanks, or click blank then click option'

A+

B*

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' instead of '+' causes incorrect mid calculation.

The middle index is found by adding low to half the distance between high and low, so use '+'.

2fill in blank

medium

Complete the code to update the low index when the target is greater than the middle element.

DSA Go

if arr[mid] < target {
    low = [1]
}

Drag options to blanks, or click blank then click option'

Amid

Bhigh

Cmid + 1

Dlow + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting low to mid causes infinite loop if mid doesn't change.

When target is greater, move low to mid + 1 to search the right half.

3fill in blank

hard

Fix the error in the loop condition to correctly run binary search while low is less than or equal to high.

DSA Go

for low [1] high {
    // binary search steps
}

Drag options to blanks, or click blank then click option'

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' misses checking when low equals high.

The loop must run while low is less than or equal to high to include all elements.

4fill in blank

hard

Fill both blanks to update the high index when the target is less than the middle element.

DSA Go

if target < arr[mid] {
    high = [1]
} else {
    low = [2]
}

Drag options to blanks, or click blank then click option'

Amid - 1

Bmid + 1

Chigh - 1

Dlow + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using high - 1 or low + 1 incorrectly for both updates.

If target is less, move high to mid - 1; else move low to mid + 1.

5fill in blank

hard

Fill all three blanks to complete the binary search function that returns the index of target or -1 if not found.

DSA Go

func binarySearch(arr []int, target int) int {
    low, high := 0, len(arr) - 1
    for low [1] high {
        mid := low + (high - low) / 2
        if arr[mid] == target {
            return [2]
        } else if arr[mid] < target {
            low = [3]
        } else {
            high = mid - 1
        }
    }
    return -1
}

Drag options to blanks, or click blank then click option'

A<=

Bmid

Cmid + 1

Dmid - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Returning low or high instead of mid; wrong loop condition; wrong low update.

The loop runs while low <= high; return mid when found; update low to mid + 1 if target is greater.

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