Find Minimum in Rotated Sorted Array in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a rotated sorted array?

A rotated sorted array is an array that was originally sorted in ascending order but then rotated (shifted) at some pivot point. For example, [4,5,6,7,0,1,2] is a rotated version of [0,1,2,4,5,6,7].

beginner

Why can't we just scan the entire rotated sorted array to find the minimum?

Scanning the entire array takes O(n) time. Using a special approach like binary search can find the minimum in O(log n) time, which is faster for large arrays.

intermediate

What is the key idea behind using binary search to find the minimum in a rotated sorted array?

We compare the middle element with the rightmost element to decide which half contains the minimum. If middle is greater than right, minimum is in the right half; else, it's in the left half.

intermediate

What is the stopping condition in the binary search for finding the minimum in a rotated sorted array?

When the search space reduces to one element, that element is the minimum.

intermediate

Show the Go code snippet to find the minimum in a rotated sorted array using binary search.

func findMin(nums []int) int { left, right := 0, len(nums)-1 for left < right { mid := left + (right-left)/2 if nums[mid] > nums[right] { left = mid + 1 } else { right = mid } } return nums[left] }

What does a rotated sorted array look like?

A [7,6,5,4,3,2,1]

B [4,5,6,7,0,1,2]

C [1,2,3,4,5,6,7]

D [2,2,2,2,2]

A rotated sorted array is a sorted array shifted at some pivot, like [4,5,6,7,0,1,2].

What is the time complexity of binary search to find the minimum in a rotated sorted array?

A O(log n)

B O(n)

C O(n log n)

D O(1)

Binary search reduces the search space by half each time, so it runs in O(log n) time.

In the binary search, if nums[mid] > nums[right], where is the minimum?

A Left half including mid

B At right

C At mid

D Right half excluding mid

If nums[mid] > nums[right], minimum must be in the right half excluding mid.

What happens when left == right in the binary search for minimum?

A We found the minimum

B We restart search

C Search continues

D We return -1

When left equals right, the search space is one element, which is the minimum.

Which of these is the correct Go function signature to find minimum in rotated sorted array?

A func findMin(nums []int) []int

B func findMin(nums int) []int

C func findMin(nums []int) int

D func findMin(nums int) int

The function takes a slice of integers and returns an integer minimum.

Explain how binary search helps find the minimum in a rotated sorted array.

Think about which half to keep based on mid and right values.

Write the steps to find the minimum in [4,5,6,7,0,1,2] using binary search.

Trace the indexes and values carefully at each step.

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