Find Minimum in Rotated Sorted Array in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find minimum element output after rotation

What is the output of the following C++ code that finds the minimum element in a rotated sorted array?

DSA C++

int findMin(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;
        }
    }
    return nums[left];
}

int main() {
    vector<int> nums = {4,5,6,7,0,1,2};
    cout << findMin(nums) << endl;
    return 0;
}

A 2

B 1

C 0

D 4

💡 Hint

Think about where the smallest element lies in a rotated sorted array.

The code uses binary search to find the minimum element. Since the array is rotated, the minimum is the only element smaller than the last element in the array segment. The code narrows down to index 4 where the value is 0.

❓ Predict Output

intermediate

Minimum element in a rotated array with duplicates

What will be the output of this code when the input array contains duplicates?

DSA C++

int findMin(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 if (nums[mid] < nums[right]) {
            right = mid;
        } else {
            right--;
        }
    }
    return nums[left];
}

int main() {
    vector<int> nums = {2,2,2,0,1,2};
    cout << findMin(nums) << endl;
    return 0;
}

A 2

B 0

C 1

D Error

💡 Hint

Duplicates require a slight modification in the binary search approach.

When duplicates exist, if nums[mid] equals nums[right], we reduce right by one to skip duplicates. The minimum element here is 0 at index 3.

🔧 Debug

advanced

Identify the error in this minimum finder code

What error will this code produce when trying to find the minimum in a rotated sorted array?

DSA C++

int findMin(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];
}

int main() {
    vector<int> nums = {3,4,5,1,2};
    cout << findMin(nums) << endl;
    return 0;
}

A Runtime error: out_of_range

B Output: 1

C Output: 3

D Infinite loop

💡 Hint

Check the loop condition and how left and right pointers move.

The loop condition allows left to go beyond right, and right is set to mid - 1 which can cause left to exceed array bounds. Accessing nums[left] after loop can cause out_of_range error.

❓ Predict Output

advanced

Output of minimum finder on a non-rotated sorted array

What is the output of this code when the input array is sorted but not rotated?

DSA C++

int findMin(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;
        }
    }
    return nums[left];
}

int main() {
    vector<int> nums = {1,2,3,4,5};
    cout << findMin(nums) << endl;
    return 0;
}

A 1

B 5

C 3

D 0

💡 Hint

If the array is not rotated, the minimum is the first element.

The code correctly returns the first element 1 as the minimum in a sorted array without rotation.

🧠 Conceptual

expert

Why binary search works for finding minimum in rotated sorted array

Which statement best explains why binary search can be used to find the minimum element in a rotated sorted array?

A Because the array elements are unique and binary search only works on unique elements.

B Because the array is rotated, the minimum element is always the last element.

C Because the minimum element is always at the middle index after rotation.

D Because the array is sorted, and rotation preserves sorted order in two parts, allowing binary search to discard half the array each step.

💡 Hint

Think about how rotation affects sorted order and how binary search uses order.

The rotated array consists of two sorted subarrays. Binary search compares middle and right elements to decide which half contains the minimum, discarding the other half 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 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