Search in Rotated Sorted Array in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the binary search loop.

DSA C++

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

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

Amid

Bright

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid instead of left in the loop condition

Using right < left instead of <=.

The binary search continues while the left index is less than or equal to the right index.

2fill in blank

medium

Complete the code to check if the middle element is the target.

DSA C++

if (nums[[1]] == target) {
    return mid;
}

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

Amid

Bright

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using left or right instead of mid to check the element.

We compare the middle element nums[mid] with the target to check for a match.

3fill in blank

hard

Fix the error in the condition to check if the left half is sorted.

DSA C++

if (nums[left] <= nums[[1]]) {
    // Left half is sorted
}

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

Aright

Bmid

Cleft

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing nums[left] with nums[right] instead of nums[mid].

We compare nums[left] with nums[mid] to check if the left half is sorted.

4fill in blank

hard

Fill both blanks to update the search boundaries correctly when the left half is sorted.

DSA C++

if (target >= nums[left] && target < nums[mid]) {
    right = [1];
} else {
    left = [2];
}

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

Amid - 1

Bmid + 1

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right updates.

Using mid instead of mid ± 1.

If the target is in the left sorted half, move right to mid - 1; otherwise, move left to mid + 1.

5fill in blank

hard

Fill all three blanks to update the search boundaries when the right half is sorted.

DSA C++

else {
    if (target > nums[mid] && target <= nums[[1]]) {
        left = [2];
    } else {
        right = [3];
    }
}

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

Aright

Bmid + 1

Cmid - 1

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Using left instead of right in comparison.

Incorrect boundary updates.

If the target is in the right sorted half, move left to mid + 1; otherwise, move right to mid - 1.

Practice

1. What is the main challenge when searching for an element in a rotated sorted array compared to a normal sorted array?

easy

A. The array is not fully sorted, so normal binary search cannot be directly applied.

B. The array contains duplicate elements which confuse the search.

C. The array size is unknown, so we cannot set search boundaries.

D. The array elements are all negative numbers.

Solution

1. Step 1: Understand normal binary search assumption

   Binary search works efficiently on fully sorted arrays by comparing mid element to target.

2. Step 2: Identify the problem with rotated arrays

   Rotated sorted arrays are sorted but shifted, so the order is broken at the rotation point, making direct binary search invalid.

3. Final Answer:

   The array is not fully sorted, so normal binary search cannot be directly applied. -> Option A

4. Quick Check:

   Rotated array breaks sorted order = need modified search [OK]

Hint: Look for the rotation point to adjust binary search [OK]

Common Mistakes:

• Assuming array is fully sorted and applying normal binary search
• Ignoring rotation and searching linearly
• Confusing duplicates with rotation

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

easy

A. if (arr[mid] == target)

B. for (int i = 0; i < n; i++)

C. while (start <= end)

D. do { ... } while (start < end)

Solution

1. Step 1: Identify loop type for binary search

   Binary search uses a while loop with condition start <= end to narrow search range.

2. Step 2: Check options for correct loop syntax

   while (start <= end) uses while (start <= end), which is standard for binary search loops.

3. Final Answer:

   while (start <= end) -> Option C

4. Quick Check:

   Binary search loop uses while (start <= end) [OK]

Hint: Binary search loops use while (start <= end) [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Using incorrect loop condition like start < end
• Confusing if statement with loop

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

int arr[] = {4,5,6,7,0,1,2};
int target = 0;
int start = 0, end = 6;
int result = -1;
while (start <= end) {
  int mid = start + (end - start) / 2;
  if (arr[mid] == target) {
    result = mid;
    break;
  }
  if (arr[start] <= arr[mid]) {
    if (target >= arr[start] && target < arr[mid])
      end = mid - 1;
    else
      start = mid + 1;
  } else {
    if (target > arr[mid] && target <= arr[end])
      start = mid + 1;
    else
      end = mid - 1;
  }
}
std::cout << result << std::endl;

medium

A. 3

B. 0

C. -1

D. 4

Solution

1. Step 1: Trace initial variables and mid calculation

   start=0, end=6, mid=3 (element 7). arr[mid] != target (0).

2. Step 2: Check which half is sorted and adjust search

   arr[start]=4 <= arr[mid]=7, left half sorted. Target 0 not in [4,7), so search right half: start=mid+1=4.

3. Step 3: Next iteration mid=5 (element 1)

   arr[mid]=1 != target. arr[start]=0 <= arr[mid]=1, left half sorted. Target 0 in [0,1), so end=mid-1=4.

4. Step 4: Next iteration mid=4 (element 0)

   arr[mid]=0 == target, result=4, break loop.

5. Final Answer:

   4 -> Option D

6. Quick Check:

   Target 0 found at index 4 [OK]

Hint: Check sorted half and adjust start/end accordingly [OK]

Common Mistakes:

• Not updating start/end correctly based on sorted half
• Confusing indices and returning wrong index
• Missing break after finding target

4. Identify the bug in the following code snippet for searching in a rotated sorted array:

int search(int arr[], int n, int target) {
  int start = 0, end = n - 1;
  while (start < end) {
    int mid = start + (end - start) / 2;
    if (arr[mid] == target) return mid;
    if (arr[start] <= arr[mid]) {
      if (target >= arr[start] && target < arr[mid])
        end = mid - 1;
      else
        start = mid + 1;
    } else {
      if (target > arr[mid] && target <= arr[end])
        start = mid + 1;
      else
        end = mid - 1;
    }
  }
  return -1;
}

medium

A. The mid calculation is incorrect and can cause overflow.

B. The loop condition should be while (start <= end) instead of while (start < end).

C. The target comparison should use > instead of >=.

D. The function does not handle empty arrays.

Solution

1. Step 1: Analyze loop condition impact

   Using while (start < end) misses checking when start == end, possibly skipping last element.

2. Step 2: Correct loop condition for complete search

   Changing to while (start <= end) ensures all elements are checked, including when start equals end.

3. Final Answer:

   The loop condition should be while (start <= end) instead of while (start < end). -> Option B

4. Quick Check:

   Loop must cover start == end to check last element [OK]

Hint: Use while (start <= end) for full binary search coverage [OK]

Common Mistakes:

• Using while (start < end) and missing last element
• Ignoring overflow in mid calculation (not in this code though)
• Incorrect comparison operators

5. Given a rotated sorted array arr = {15, 18, 2, 3, 6, 12}, write a C++ function to find the index of the minimum element. What is the output of your function?

hard

A. 2

B. 0

C. 5

D. 3

Solution

1. Step 1: Understand the minimum element in rotated array

   The minimum element is the rotation point where order breaks. Here, 2 at index 2 is smallest.

2. Step 2: Use binary search to find minimum

   Check mid and compare with end to decide which half contains minimum. Eventually, index 2 is found.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Minimum element 2 is at index 2 [OK]

Hint: Compare mid element with end to find rotation point [OK]

Common Mistakes:

• Returning first element index instead of minimum
• Confusing maximum with minimum element
• Not handling rotation correctly