Find Minimum in Rotated Sorted Array in DSA Python - 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].

intermediate

Why can we use binary search to find the minimum in a rotated sorted array?

Because the array is mostly sorted except for the rotation point, we can use binary search to efficiently find the smallest element by comparing middle elements with the ends to decide which half to search next.

beginner

In the rotated sorted array [4,5,6,7,0,1,2], what is the minimum element and its index?

The minimum element is 0, and its index is 4.

intermediate

What is the time complexity of finding the minimum in a rotated sorted array using binary search?

The time complexity is O(log n) because binary search divides the search space in half each step.

intermediate

What condition tells us that the middle element is greater than the rightmost element in the rotated sorted array?

If the middle element is greater than the rightmost element, it means the minimum is in the right half of the array because the rotation point is there.

What does a rotated sorted array look like?

A An array sorted in descending order

B An array sorted in ascending order but shifted at some pivot

C An array with all elements equal

D A random unsorted array

A rotated sorted array is originally sorted ascending but rotated at a pivot point.

Which algorithm is best to find the minimum in a rotated sorted array efficiently?

A Binary search

B Selection sort

C Bubble sort

D Linear search

Binary search efficiently finds the minimum in O(log n) time by halving the search space.

If the middle element is less than the rightmost element, where is the minimum likely located?

A In the left half including middle

B In the right half excluding middle

C At the middle only

D At the rightmost element only

If middle < right, minimum is in left half including middle.

What is the minimum element in [3,4,5,1,2]?

A 2

B 3

C 1

D 5

1 is the smallest element after rotation.

What is the time complexity of a linear search for minimum in an array of size n?

A O(1)

B O(log n)

C O(n log n)

D O(n)

Linear search checks each element, so it takes O(n) time.

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

Think about how the rotation affects the order and how to use comparisons to find the pivot.

Describe the characteristics of a rotated sorted array and how it differs from a normal sorted array.

Imagine a sorted list shifted so the smallest element is not at the start.

Practice

1. What is the main goal when finding the minimum in a rotated sorted array like [4, 5, 6, 7, 0, 1, 2]?

easy

A. To find the smallest number where the rotation happened

B. To find the largest number in the array

C. To sort the array again

D. To count how many times the array was rotated

Solution

1. Step 1: Understand the rotated sorted array

   A rotated sorted array is originally sorted but then shifted. The smallest number is where the rotation happened.

2. Step 2: Identify the minimum element

   The minimum element is the smallest number, marking the rotation point.

3. Final Answer:

   To find the smallest number where the rotation happened -> Option A

4. Quick Check:

   Minimum = smallest number at rotation [OK]

Hint: Minimum is the smallest number where order breaks [OK]

Common Mistakes:

• Looking for the largest number instead
• Trying to sort the array again
• Counting rotations instead of finding minimum

2. Which of the following Python code snippets correctly initializes variables to find the minimum in a rotated sorted array nums using binary search?

easy

A. start, end = len(nums), 0

B. start = 1; end = len(nums)

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

D. start = -1; end = len(nums) + 1

Solution

1. Step 1: Understand binary search initialization

   Binary search uses start at 0 and end at last index (length - 1).

2. Step 2: Check each option

   start, end = 0, len(nums) - 1 correctly sets start=0 and end=len(nums)-1. Others have invalid ranges.

3. Final Answer:

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

4. Quick Check:

   Binary search range = start=0, end=length-1 [OK]

Hint: Start at 0, end at last index for binary search [OK]

Common Mistakes:

• Setting start or end outside valid index range
• Using length instead of length - 1 for end
• Starting from 1 instead of 0

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

def find_min(nums):
    start, end = 0, len(nums) - 1
    while start < end:
        mid = (start + end) // 2
        if nums[mid] > nums[end]:
            start = mid + 1
        else:
            end = mid
    return nums[start]

print(find_min([3, 4, 5, 1, 2]))

medium

A. 2

B. 3

C. 5

D. 1

Solution

1. Step 1: Trace the binary search steps

   Initial array: [3,4,5,1,2], start=0, end=4 - mid=2, nums[mid]=5, nums[end]=2, 5>2 so start=mid+1=3 - now start=3, end=4 - mid=3, nums[mid]=1, nums[end]=2, 1<=2 so end=mid=3 - start=3, end=3, loop ends

2. Step 2: Return minimum element

   nums[start]=nums[3]=1 is the minimum.

3. Final Answer:

   1 -> Option D

4. Quick Check:

   Minimum found = 1 [OK]

Hint: Compare mid and end to adjust search [OK]

Common Mistakes:

• Returning nums[mid] instead of nums[start]
• Not updating start or end correctly
• Confusing > and < in condition

4. The following code is intended to find the minimum in a rotated sorted array, but it has a bug. What is the bug?

def find_min(nums):
    start, end = 0, len(nums) - 1
    while start <= end:
        mid = (start + end) // 2
        if nums[mid] > nums[end]:
            start = mid + 1
        else:
            end = mid
    return nums[start]

medium

A. The loop condition should be start < end, not start <= end

B. The return statement should be nums[end], not nums[start]

C. The mid calculation is incorrect

D. The if condition should compare nums[mid] with nums[start]

Solution

1. Step 1: Analyze the loop condition

   The loop uses start <= end, which can cause infinite loop or index errors because start can pass end.

2. Step 2: Correct loop condition for binary search

   Using start < end ensures the loop stops when start meets end, correctly locating minimum.

3. Final Answer:

   The loop condition should be start < end, not start <= end -> Option A

4. Quick Check:

   Loop condition must be start < end [OK]

Hint: Use start < end for binary search loops [OK]

Common Mistakes:

• Using <= causes infinite loops or wrong indexes
• Changing mid calculation unnecessarily
• Comparing with wrong array element

5. Given a rotated sorted array that may contain duplicates, such as [2, 2, 2, 0, 1, 2], which modification to the binary search approach helps correctly find the minimum?

hard

A. Sort the array first before searching

B. When nums[mid] == nums[end], reduce end by 1 to skip duplicate

C. Always move start to mid + 1 when nums[mid] >= nums[end]

D. Return nums[0] directly as minimum

Solution

1. Step 1: Understand duplicates affect comparison

   When nums[mid] equals nums[end], we cannot decide which side minimum is on because duplicates hide order.

2. Step 2: Skip duplicate by reducing end

   Reducing end by 1 removes one duplicate and narrows search without missing minimum.

3. Final Answer:

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

4. Quick Check:

   Handle duplicates by shrinking end [OK]

Hint: If mid equals end, reduce end by 1 to handle duplicates [OK]

Common Mistakes:

• Assuming no duplicates and using normal binary search
• Sorting array first wastes time
• Returning first element without checking