Find Minimum in Rotated Sorted Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the first element of the array.

DSA C++

int findMin(vector<int>& nums) {
    return nums[[1]];
}

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

Anums.size()

B1

Cnums.size() - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 1 instead of 0

Using nums.size() which is out of bounds

The minimum element in a non-rotated sorted array is the first element at index 0.

2fill in blank

medium

Complete the code to find the middle index in binary search.

DSA C++

int left = 0, right = nums.size() - 1;
while (left < right) {
    int mid = (left [1] right) / 2;
    // rest of code
}

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A+

B+ +

Attempts:

3 left

💡 Hint

Common Mistakes

Using only left or right

Using bit shifts incorrectly

The middle index is calculated as (left + right) / 2.

3fill in blank

hard

Fix the error in the condition to check if mid element is greater than the right element.

DSA C++

if (nums[mid] [1] nums[right]) {
    left = mid + 1;
} else {
    right = mid;
}

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A<

B==

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of >

Using == which is incorrect here

If nums[mid] is greater than nums[right], the minimum is in the right half, so move left to mid + 1.

4fill in blank

hard

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

DSA C++

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

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Aleft

Bright

Cmid

Dnums.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid in the condition

Using nums.size() which is out of bounds

The loop continues while left is less than right to narrow down the search space.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each number to its square if the number is less than 5.

DSA C++

auto squares = std::unordered_map<int, int>{{
    { [1], [2] } for (int num : nums) if (num [3] 5)
}};

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

Anum

Bnum * num

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<'

Using num instead of num * num for value

This comprehension maps each number to its square only if the number is less than 5.

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