Why Binary Search and What Sorted Order Gives You in DSA Go - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle index in binary search.

DSA Go

mid := (low + [1]) / 2

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

Alow

Bhigh

Cmid

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Using low twice instead of high

Using length of array directly

In binary search, the middle index is calculated by adding low and high indices and dividing by 2.

2fill in blank

medium

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

DSA Go

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

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

Alow

Bhigh

Cmid

Dtarget

Attempts:

3 left

💡 Hint

Common Mistakes

Updating high instead of low

Not adding 1 to mid

If the target is greater than the middle element, we move the low index to mid + 1 to search the right half.

3fill in blank

hard

Fix the error in the loop condition to correctly run binary search.

DSA Go

for low [1] high {
    mid := (low + high) / 2
    // rest of code
}

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

A<=

B<

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using only less than which misses last element

Using greater than

The loop should continue while low is less than or equal to high to cover all elements.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

DSA Go

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

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

A>

Blen(word)

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than instead of greater than

Storing word instead of length

We check if word length is greater than 3, then store the length in the map.

5fill in blank

hard

Fill all three blanks to create a filtered map with uppercase keys and values greater than 0.

DSA Go

filtered := map[string]int{}
for k, v := range data {
    if v [1] 0 {
        filtered[[2]] = v
    }
}

// Convert keys to uppercase
for k := range filtered {
    newKey := [3]
    filtered[newKey] = filtered[k]
    delete(filtered, k)
}

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

A>

Bk

Cstrings.ToUpper(k)

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' instead of '>'

Using value instead of key for map assignment

Not converting key to uppercase

We filter values greater than 0, keep original keys, then convert keys to uppercase.

Practice

1. Why is binary search only applicable on a sorted list?

easy

A. Because it relies on dividing the list based on order to find elements quickly

B. Because it works only on lists with unique elements

C. Because it requires the list to be reversed

D. Because it compares every element one by one

Solution

1. Step 1: Understand binary search logic

   Binary search splits the list into halves and decides which half to search next based on order.

2. Step 2: Importance of sorted order

   Without sorted order, we cannot decide which half to discard, so binary search fails.

3. Final Answer:

   Because it relies on dividing the list based on order to find elements quickly -> Option A

4. Quick Check:

   Sorted list needed for binary search = Because it relies on dividing the list based on order to find elements quickly [OK]

Hint: Binary search needs order to skip half the list each step [OK]

Common Mistakes:

• Thinking binary search works on unsorted lists
• Believing binary search compares all elements
• Assuming binary search needs unique elements

2. Which of the following Go code snippets correctly performs binary search on a sorted slice arr to find target?

easy

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

B. for i := 0; i < len(arr); i++ {
if arr[i] == target {
return i
}
}
return -1

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

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

Solution

1. Step 1: Check binary search loop and boundaries

   left, right := 0, len(arr)-1 for left <= right { mid := (left + right) / 2 if arr[mid] == target { return mid } else if arr[mid] < target { left = mid + 1 } else { right = mid - 1 } } return -1 correctly initializes left and right, uses <=, and updates left and right properly.

2. Step 2: Identify errors in other options

   for i := 0; i < len(arr); i++ { if arr[i] == target { return i } } return -1 is linear search, not binary. left, right := 0, len(arr) for left < right { mid := (left + right) / 2 if arr[mid] == target { return mid } else if arr[mid] > target { left = mid + 1 } else { right = mid - 1 } } return -1 has wrong boundary and updates. left, right := 0, len(arr)-1 for left <= right { mid := (left + right) / 2 if arr[mid] == target { return mid } else if arr[mid] > target { right = mid - 1 } else { left = mid - 1 } } return -1 incorrectly updates left pointer.

3. Final Answer:

   Option A code correctly implements binary search -> Option A

4. Quick Check:

   Correct binary search syntax = left, right := 0, len(arr)-1 for left <= right { mid := (left + right) / 2 if arr[mid] == target { return mid } else if arr[mid] < target { left = mid + 1 } else { right = mid - 1 } } return -1 [OK]

Hint: Binary search updates left/right pointers correctly [OK]

Common Mistakes:

• Using linear search instead of binary search
• Wrong loop conditions like < instead of <=
• Incorrectly updating left or right pointers

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
for left <= right {
  mid := (left + right) / 2
  if arr[mid] == target {
    fmt.Println(mid)
    break
  } else if arr[mid] < target {
    left = mid + 1
  } else {
    right = mid - 1
  }
}

medium

A. 0

B. 3

C. 1

D. 2

Solution

1. Step 1: Trace binary search steps

   Initial left=0, right=4, mid=2 (value 6). arr[mid] == target, so print 2 and break.

2. Step 2: Confirm output

   The printed index is 2, which is the position of 6 in the slice.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Index of 6 in arr = 2 [OK]

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

Common Mistakes:

• Confusing index with value
• Miscomputing mid index
• Ignoring break after finding target

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

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

medium

A. The array must be reversed before binary search

B. Loop condition should be left <= right instead of left < right

C. left and right pointers are updated incorrectly

D. mid calculation should be mid := (left + right + 1) / 2

Solution

1. Step 1: Analyze loop condition

   The loop uses left < right, which misses the case when left == right, possibly skipping the last element.

2. Step 2: Correct loop condition

   Changing to left <= right ensures all elements are checked, fixing the bug.

3. Final Answer:

   Loop condition should be left <= right instead of left < right -> Option B

4. Quick Check:

   Loop must cover all indices = Loop condition should be left <= right instead of left < right [OK]

Hint: Use <= in loop to include last element check [OK]

Common Mistakes:

• Using < instead of <= in loop condition
• Changing mid calculation unnecessarily
• Thinking array must be reversed

5. Given a sorted slice arr := []int{1, 3, 5, 7, 9, 11, 13}, you want to find the smallest element greater than or equal to target = 8. Which modification to binary search helps achieve this?

hard

A. Use linear search instead of binary search

B. When arr[mid] > target, move left to mid + 1; else move right to mid - 1

C. When arr[mid] < target, move left to mid + 1; else move right to mid - 1 and track mid as potential answer

D. Reverse the array and perform binary search

Solution

1. Step 1: Understand the problem

   We want the smallest element >= target, so we track candidates when arr[mid] >= target.

2. Step 2: Modify binary search accordingly

   If arr[mid] < target, move left to mid + 1; else move right to mid - 1 and save mid as potential answer.

3. Final Answer:

   When arr[mid] < target, move left to mid + 1; else move right to mid - 1 and track mid as potential answer -> Option C

4. Quick Check:

   Track candidate when arr[mid] >= target = When arr[mid] < target, move left to mid + 1; else move right to mid - 1 and track mid as potential answer [OK]

Hint: Track mid when arr[mid] >= target, adjust pointers accordingly [OK]

Common Mistakes:

• Not tracking potential answer index
• Moving pointers incorrectly
• Using linear search instead of binary search