Binary Search vs Linear Search Real Cost Difference in DSA Go - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform a linear search that returns the index of target or -1 if not found.

DSA Go

func linearSearch(arr []int, target int) int {
    for i := 0; i < len(arr); i++ {
        if arr[i] == [1] {
            return i
        }
    }
    return -1
}

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

Alen(arr)

Bi

Carr

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[i] with i instead of target

Using len(arr) instead of target in comparison

The code compares each element arr[i] with the target value to find a match.

2fill in blank

medium

Complete the code to perform a binary search on a sorted array to find the target index or -1 if not found.

DSA Go

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

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

Amid

Btarget

Clow

Dhigh

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing arr[mid] with mid instead of target

Using low or high instead of target in comparison

The binary search compares the middle element arr[mid] with the target to decide the search direction.

3fill in blank

hard

Fix the error in the binary search code: replace the incorrect operator to correctly update the high index.

DSA Go

if arr[mid] > target {
    [1] = mid - 1
}

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

Alow

Btarget

Chigh

Dmid

Attempts:

3 left

💡 Hint

Common Mistakes

Updating low instead of high

Assigning mid or target instead of high

When arr[mid] is greater than target, we reduce the high index to mid - 1 to search the left half.

4fill in blank

hard

Fill both blanks to create a loop that runs while low is less than or equal to high and calculates mid correctly.

DSA Go

for [1] <= [2] {
    mid := (low + high) / 2
    // search logic
}

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

Alow

Bhigh

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid or target in the loop condition

Reversing low and high in the condition

The loop continues while low is less than or equal to high to cover all elements in the search range.

5fill in blank

hard

Fill all three blanks to create a map comprehension that stores word lengths for words longer than 3 characters.

DSA Go

lengths := map[string]int{}
for _, word := range words {
    if len(word) [1] 3 {
        lengths[[2]] = [3]
    }
}

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

A>

Bword

Clen(word)

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than operator instead of greater than

Using len(word) as key or word as value

We check if the word length is greater than 3, then store the word as key and its length as value.

Practice

1. Which search algorithm is generally faster for finding an element in a sorted list of 1,000,000 items?

easy

A. Neither works on sorted lists

B. Linear Search

C. Binary Search

D. Both are equally fast

Solution

1. Step 1: Understand the search algorithms

   Binary Search divides the list in half repeatedly, while Linear Search checks each item one by one.

2. Step 2: Compare efficiency on large sorted lists

   Binary Search takes about log2(1,000,000) ≈ 20 steps, Linear Search may take up to 1,000,000 steps.

3. Final Answer:

   Binary Search -> Option C

4. Quick Check:

   Binary Search is faster on sorted large lists [OK]

Hint: Binary Search halves list; Linear checks all items [OK]

Common Mistakes:

• Thinking Linear Search is faster on sorted lists
• Confusing sorted vs unsorted list requirements
• Assuming both have same speed

2. Which of the following is the correct Go syntax to perform a linear search on a slice named arr for a value target?

easy

A. for i, v := range arr { if v == target { return v } }

B. for i := 0; i < len(arr); i++ { if arr[i] == target { return i } }

C. for v in arr { if v == target { return i } }

D. while i < len(arr) { if arr[i] == target { return i } i++ }

Solution

1. Step 1: Review Go loop syntax for slices

   Go uses for loops with index and length or range loops; for v in arr { if v == target { return i } } uses invalid syntax, while i < len(arr) { if arr[i] == target { return i } i++ } uses while which Go does not have.

2. Step 2: Identify correct syntax for linear search

   for i := 0; i < len(arr); i++ { if arr[i] == target { return i } } uses a for loop with index and length, checking each element correctly.

3. Final Answer:

   for i := 0; i < len(arr); i++ { if arr[i] == target { return i } } -> Option B

4. Quick Check:

   Go uses for loops, not while [OK]

Hint: Go uses for loops, not while loops [OK]

Common Mistakes:

• Using Python or other language syntax in Go
• Using while loops which Go does not support
• Incorrect range loop syntax

3. Given the following Go code snippet, what will be printed?

func linearSearch(arr []int, target int) int {
    for i := 0; i < len(arr); i++ {
        if arr[i] == target {
            return i
        }
    }
    return -1
}

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

medium

A. 2

B. 3

C. -1

D. 0

Solution

1. Step 1: Trace the linear search function

   The function checks each element in order: arr[0]=3, arr[1]=5, arr[2]=7 matches target 7.

2. Step 2: Return index of first match

   It returns index 2 immediately when it finds 7.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   7 is at index 2 in arr [OK]

Hint: Linear search returns first matching index [OK]

Common Mistakes:

• Confusing index with value
• Assuming zero-based index is one-based
• Returning -1 incorrectly

4. The following Go code attempts to perform binary search on a sorted slice but does not work correctly. What is the main bug?

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

medium

A. low should start at 1 instead of 0

B. mid calculation should use float division

C. The loop condition should be low < high

D. high should be initialized to len(arr) - 1

Solution

1. Step 1: Check initialization of high index

   In Go, slice indices go from 0 to len(arr)-1, so high should be len(arr)-1, not len(arr).

2. Step 2: Understand impact on loop and mid calculation

   Using len(arr) causes out-of-range access when mid equals len(arr), causing runtime error.

3. Final Answer:

   high should be initialized to len(arr) - 1 -> Option D

4. Quick Check:

   Slice max index = len(arr)-1 [OK]

Hint: High index = length minus one for slices [OK]

Common Mistakes:

• Setting high to len(arr) causes out-of-range error
• Using float division for mid index
• Incorrect loop condition

5. You have a sorted slice of 1,000,000 integers and want to find if the number 123456 exists. You can use either linear or binary search. Considering real cost difference, which approach is best and why?

hard

A. Use binary search because it reduces comparisons drastically

B. Use linear search because it is simpler to implement

C. Use linear search because it uses less memory

D. Use binary search only if the slice is unsorted

Solution

1. Step 1: Analyze the data and search requirements

   The slice is sorted, so binary search can be applied efficiently.

2. Step 2: Compare real cost of linear vs binary search

   Linear search may check up to 1,000,000 elements; binary search checks about 20 comparisons, saving time.

3. Final Answer:

   Use binary search because it reduces comparisons drastically -> Option A

4. Quick Check:

   Binary search best for sorted large data [OK]

Hint: Binary search cuts steps to log2(n) [OK]

Common Mistakes:

• Choosing linear search for large sorted data
• Confusing memory use with search speed
• Thinking binary search needs unsorted data