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
}

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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
}

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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
}

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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.