Find Minimum in Rotated Sorted Array in DSA C++ - 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 unknown to you. 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, which is slow for large arrays. Using a smart approach like binary search can find the minimum in O(log n) time, which is much faster.

intermediate

In the binary search approach to find the minimum in a rotated sorted array, what does it mean 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 to the right of mid.

intermediate

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

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

intermediate

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

Start with low=0, high=6. mid=3 (value=7). Since 7 > 2 (rightmost), move low to mid+1=4. Now low=4, high=6. mid=5 (value=1). Since 1 < 2, move high to mid=5. Now low=4, high=5. mid=4 (value=0). Since 0 < 1, move high to mid=4. Now low=4, high=4. Stop. Minimum is 0 at index 4.

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

A O(n log n)

B O(log n)

C O(n)

D O(1)

Binary search divides the search space in half each time, leading to O(log n) time complexity.

If the middle element is less than the rightmost element in the rotated sorted array, where is the minimum located?

A In the left half including mid

B In the right half excluding mid

C At the middle element only

D Cannot determine

If mid element is less than rightmost, the minimum lies in the left half including mid.

What does it mean if the array is not rotated?

A The array is empty

B The last element is the minimum

C The middle element is the minimum

D The first element is the minimum

If the array is not rotated, it is fully sorted, so the first element is the smallest.

Which of these arrays is a rotated sorted array?

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

B [1,2,3,4,5]

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

D [2,1,3,4,5]

Array [3,4,5,1,2] is a rotated version of sorted array [1,2,3,4,5].

What is the first step in the binary search approach to find the minimum in a rotated sorted array?

A Compare middle element with leftmost element

B Scan the entire array

C Compare middle element with rightmost element

D Sort the array

We compare the middle element with the rightmost element to decide which half to search next.

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 difference between a rotated sorted array and a fully sorted array and how it affects finding the minimum.

Consider where the smallest number is located in each case.

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 the minimum

B. Sort the array again and pick the first element

C. Traverse the array linearly and pick the smallest element

D. Use binary search to find the point where the 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 pivot where the order breaks, which is the minimum element.

3. Final Answer:

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

4. Quick Check:

   Binary search finds minimum in rotated array [OK]

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

Common Mistakes:

• Trying to sort the array again wastes time
• Linear search is slower than binary search
• Using extra space unnecessarily

2. Which of the following is the correct syntax to start a binary search loop in C++ for finding minimum in a rotated sorted array named nums?

easy

A. while (left <= right) { int mid = left + (right - left) / 2; }

B. while (left < right) { int mid = (left + right) / 2; }

C. for (int left = 0, right = nums.size() - 1; left <= right; )

D. for (int i = 0; i < nums.size(); i++)

Solution

1. Step 1: Understand binary search loop conditions

   For finding minimum in a rotated sorted array, the binary search loop typically uses while (left < right) to avoid infinite loops and correctly narrow down the search space.

2. Step 2: Check correct mid calculation and loop condition

   while (left < right) { int mid = (left + right) / 2; } is the common and correct approach for this problem.

3. Final Answer:

   while (left < right) { int mid = (left + right) / 2; } -> Option B

4. Quick Check:

   Correct binary search loop syntax for this problem [OK]

Hint: Use while with left < right and mid calculation [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Calculating mid as left + (right - left) / 2 with wrong loop condition
• Using wrong loop condition like left <= right causing infinite loop

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

std::vector<int> nums = {4,5,6,7,0,1,2};
int left = 0, right = nums.size() - 1;
while (left < right) {
  int mid = left + (right - left) / 2;
  if (nums[mid] > nums[right]) left = mid + 1;
  else right = mid;
}
std::cout << nums[left] << std::endl;

medium

A. 0

B. 4

C. 7

D. 2

Solution

1. Step 1: Understand the binary search logic on rotated array

   The code compares nums[mid] with nums[right] to decide which half to keep.

2. Step 2: Trace the loop to find minimum

   Since nums[mid] > nums[right], left moves to mid+1 until left == right at index of minimum (0).

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Minimum element found is 0 [OK]

Hint: Compare mid and right to narrow search for minimum [OK]

Common Mistakes:

• Confusing mid and right comparison direction
• Printing nums[right] instead of nums[left]
• Not updating left or right correctly

4. Identify the error in the following code snippet that tries to find the minimum in a rotated sorted array:

int findMin(std::vector<int>& nums) {
  int left = 0, right = nums.size() - 1;
  while (left <= right) {
    int mid = left + (right - left) / 2;
    if (nums[mid] > nums[right]) left = mid + 1;
    else right = mid - 1;
  }
  return nums[left];
}

medium

A. The else block should set right = mid, not right = mid - 1

B. The loop condition should be left < right, not left <= right

C. The mid calculation causes overflow

D. The return statement should return nums[right], not nums[left]

Solution

1. Step 1: Analyze the binary search update rules

   When nums[mid] <= nums[right], the minimum is at mid or left side, so right should be set to mid.

2. Step 2: Identify incorrect update causing skipping minimum

   Setting right = mid - 1 skips mid, possibly missing the minimum element.

3. Final Answer:

   The else block should set right = mid, not right = mid - 1 -> Option A

4. Quick Check:

   Correct right update avoids missing minimum [OK]

Hint: Set right = mid to include mid in search [OK]

Common Mistakes:

• Using left <= right causes infinite loop
• Returning nums[left] is correct
• Mid calculation is safe here

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

hard

A. If nums[mid] == nums[right], return nums[mid] immediately

B. If nums[mid] == nums[right], increase left by one to skip duplicate

C. If nums[mid] == nums[right], reduce right by one to skip duplicate

D. Duplicates do not affect the binary search logic

Solution

1. Step 1: Understand impact of duplicates on binary search

   Duplicates can cause ambiguity when nums[mid] == nums[right], making it unclear which side to choose.

2. Step 2: Modify search by reducing right to skip duplicate

   Reducing right by one helps shrink the search space safely without missing the minimum.

3. Final Answer:

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

4. Quick Check:

   Handle duplicates by shrinking right pointer [OK]

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

Common Mistakes:

• Returning immediately on equality misses minimum
• Increasing left instead of reducing right causes errors
• Ignoring duplicates leads to wrong answers