Find Minimum in Rotated Sorted Array in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the first element as minimum when the array is not rotated.

DSA Go

if nums[0] [1] nums[len(nums)-1] {
    return nums[0]
}

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

A==

B>

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes wrong rotation check.

Using '==' or '!=' does not correctly detect rotation.

If the first element is greater than the last, the array is rotated. Otherwise, the first element is the minimum.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA Go

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

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

Alow

Bhigh

Clen(nums)

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using (low + high) / 2 can cause overflow in some languages.

Using 0 or len(nums) as middle index is incorrect.

To avoid overflow, middle index is calculated as low + (high - low) / 2.

3fill in blank

hard

Fix the error in the condition to decide which half to search next.

DSA Go

if nums[mid] [1] nums[high] {
    low = mid + 1
} else {
    high = mid
}

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

A!=

B<

C==

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<' reverses the search direction.

Using '==' or '!=' causes incorrect logic.

If middle element is greater than the last element, minimum is in the right half.

4fill in blank

hard

Fill both blanks to complete the binary search loop and return statement.

DSA Go

for low [1] high {
    mid := low + (high - low) / 2
    if nums[mid] > nums[high] {
        low = mid + 1
    } else {
        high = mid
    }
}
return nums[2]

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

A<=

B<

C[low]

D[high]

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' instead of '<' causes infinite loop.

Returning nums[high] instead of nums[low] is incorrect.

The loop runs while low is less than high. After the loop, nums[low] is the minimum.

5fill in blank

hard

Fill all three blanks to create a function that finds the minimum in a rotated sorted array.

DSA Go

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

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

A>

B<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators in conditions.

Incorrect loop condition causing infinite loop or missed elements.

The function checks if array is rotated, then uses binary search with correct comparisons to find minimum.

Practice

1. What is the main idea behind finding the minimum element in a rotated sorted array?

easy

A. Use a hash map to store elements and find minimum

B. Sort the array again to find the minimum

C. Traverse the array linearly to find the smallest element

D. Use binary search to find the point where order breaks

Solution

1. Step 1: Understand the rotated sorted array property

   A rotated sorted array is sorted but rotated at some pivot point, causing a break in order.

2. Step 2: Use binary search to find the break point

   Binary search helps find the point where the order breaks, which is where the minimum element lies.

3. Final Answer:

   Use binary search to find the point where order breaks -> Option D

4. Quick Check:

   Binary search finds minimum efficiently [OK]

Hint: Look for the pivot where order breaks using binary search [OK]

Common Mistakes:

• Trying to sort the array again wastes time
• Using linear search ignores the sorted property
• Using extra space unnecessarily

2. Which Go code snippet correctly initializes variables for binary search to find minimum in a rotated sorted array nums?

easy

A. start, end := -1, len(nums)-1

B. start, end := 0, len(nums)-1

C. start, end := 1, len(nums)

D. start, end := 0, len(nums)

Solution

1. Step 1: Understand array indexing in Go

   Arrays are zero-indexed, so start should be 0 and end should be last index (length - 1).

2. Step 2: Check each option for correct initialization

   start, end := 0, len(nums)-1 correctly sets start=0 and end=len(nums)-1, which is valid for binary search.

3. Final Answer:

   start, end := 0, len(nums)-1 -> Option B

4. Quick Check:

   Correct index range initialization [OK]

Hint: Start at 0 and end at length minus one for array indices [OK]

Common Mistakes:

• Using length as end index causes out-of-range errors
• Starting at 1 misses the first element
• Negative start index is invalid

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

nums := []int{4,5,6,7,0,1,2}
start, end := 0, len(nums)-1
for start < end {
  mid := (start + end) / 2
  if nums[mid] > nums[end] {
    start = mid + 1
  } else {
    end = mid
  }
}
fmt.Println(nums[start])

medium

A. 0

B. 4

C. 7

D. 2

Solution

1. Step 1: Trace the binary search loop

   Initial start=0, end=6. Mid=(0+6)/2=3, nums[3]=7 > nums[6]=2, so start=4.

2. Step 2: Continue loop until start meets end

   Now start=4, end=6. Mid=(4+6)/2=5, nums[5]=1 <= nums[6]=2, so end=5. Next mid=(4+5)/2=4, nums[4]=0 <= nums[5]=1, so end=4. Loop ends when start=4, end=4.

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Minimum element found at index 4 is 0 [OK]

Hint: Compare mid with end to decide search side [OK]

Common Mistakes:

• Confusing mid calculation causing infinite loop
• Incorrectly updating start or end pointers
• Printing nums[end] instead of nums[start]

4. Identify the bug in this Go code to find minimum in rotated sorted array nums:

start, end := 0, len(nums)-1
for start <= end {
  mid := (start + end) / 2
  if nums[mid] > nums[end] {
    start = mid
  } else {
    end = mid - 1
  }
}
fmt.Println(nums[start])

medium

A. All of the above

B. Loop condition should be start < end, not <=

C. end should be set to mid, not mid - 1

D. start should be set to mid + 1, not mid

Solution

1. Step 1: Analyze loop condition and pointer updates

   The loop uses start <= end which can cause infinite loop when start == end. Also, start = mid does not move start forward properly, and end = mid - 1 skips the mid element incorrectly.

2. Step 2: Correct the loop and pointer updates

   Loop should be start < end. When nums[mid] > nums[end], start = mid + 1 to move forward. Else, end = mid to include mid in next search.

3. Final Answer:

   All of the above -> Option A

4. Quick Check:

   Proper loop and pointer updates avoid infinite loops [OK]

Hint: Use start < end and adjust pointers by +1 or no change [OK]

Common Mistakes:

• Using <= causes infinite loops
• Not moving start pointer forward correctly
• Skipping mid element by subtracting 1 from end

5. Given a rotated sorted array nums that may contain duplicates, which modification is needed in the binary search approach to find the minimum element?

hard

A. Ignore duplicates and apply normal binary search

B. If nums[mid] == nums[end], increase start by 1 to skip duplicate

C. If nums[mid] == nums[end], reduce end by 1 to skip duplicate

D. Sort the array first to handle duplicates

Solution

1. Step 1: Understand duplicates affect binary search

   Duplicates can cause nums[mid] == nums[end], making it unclear which side to search.

2. Step 2: Modify binary search to handle duplicates

   When nums[mid] == nums[end], safely reduce end by 1 to skip duplicate without missing minimum.

3. Final Answer:

   If nums[mid] == nums[end], reduce end by 1 to skip duplicate -> Option C

4. Quick Check:

   Handle duplicates by shrinking search space carefully [OK]

Hint: When equal, reduce end to skip duplicates safely [OK]

Common Mistakes:

• Ignoring duplicates causes wrong results
• Increasing start instead of reducing end misses minimum
• Sorting wastes time and is unnecessary