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 with the correct initial left index.

DSA C#

int left = [1];

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

A0

B1

C-1

Dnums.Length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 skips the first element.

Using -1 causes index errors.

Using nums.Length is out of bounds.

The search starts from the first index, which is 0.

2fill in blank

medium

Complete the code to calculate the middle index correctly.

DSA C#

int mid = left + ([1] - left) / 2;

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

A0

Bnums.Length

Cmid

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums.Length causes wrong mid calculation.

Using mid in calculation causes infinite loop.

Using 0 ignores the right boundary.

We calculate mid between left and right to avoid overflow.

3fill in blank

hard

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

DSA C#

if (nums[left] [1] nums[mid])

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using > reverses the logic.

Using <= or >= includes equality which may cause errors.

The left half is sorted if nums[left] is less than nums[mid].

4fill in blank

hard

Fill both blanks to update the search range when the target is in the left sorted half.

DSA C#

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

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong inequality signs reverses the condition.

Using <= instead of < for target < nums[mid] may cause off-by-one errors.

Target must be less than nums[mid] and greater or equal to nums[left] to be in left half.

5fill in blank

hard

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

DSA C#

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

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

A<=

B>

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - moves right pointer in wrong direction.

Using wrong inequality signs causes wrong search range.

Target is between mid and right, so left moves up and right moves down by 1.

Practice

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

easy

A. The array is sorted in descending order instead of ascending

B. The array contains duplicate elements that confuse the search

C. The array is split into two sorted parts, making normal binary search invalid directly

D. The array size is always odd, which affects indexing

Solution

1. Step 1: Understand the array structure

   A rotated sorted array is created by taking a sorted array and rotating it at some pivot. This splits it into two sorted parts.

2. Step 2: Why normal binary search fails

   Normal binary search assumes the entire array is sorted. Here, because of rotation, one half is sorted and the other may not be, so direct binary search won't work.

3. Final Answer:

   The array is split into two sorted parts, making normal binary search invalid directly -> Option C

4. Quick Check:

   Rotation splits array = challenge for binary search [OK]

Hint: Look for the pivot splitting the array into two sorted parts [OK]

Common Mistakes:

• Assuming the whole array is sorted
• Ignoring the pivot point
• Confusing rotation with reversal

2. Which of the following C# code snippets correctly initializes variables for binary search in a rotated sorted array?

easy

A. int left = -1, right = nums.Length - 1;

B. int left = 1, right = nums.Length;

C. int left = 0, right = nums.Length;

D. int left = 0, right = nums.Length - 1;

Solution

1. Step 1: Understand array indexing in C#

   Arrays are zero-indexed, so the first element is at 0 and the last at Length - 1.

2. Step 2: Correct initialization for binary search

   Binary search starts with left at 0 and right at last index (Length - 1). This covers the entire array.

3. Final Answer:

   int left = 0, right = nums.Length - 1; -> Option D

4. Quick Check:

   Zero-based indexing means right = Length - 1 [OK]

Hint: Start left at 0 and right at last index (Length - 1) [OK]

Common Mistakes:

• Setting right to Length instead of Length - 1
• Starting left at 1 instead of 0
• Using negative indices

3. What will be the output of the following C# code snippet searching for target 3 in rotated array int[] nums = {4,5,6,7,0,1,2};?

int left = 0, right = nums.Length - 1;
int target = 3;
while (left <= right) {
    int mid = left + (right - left) / 2;
    if (nums[mid] == target) return mid;
    if (nums[left] <= nums[mid]) {
        if (nums[left] <= target && target < nums[mid]) right = mid - 1;
        else left = mid + 1;
    } else {
        if (nums[mid] < target && target <= nums[right]) left = mid + 1;
        else right = mid - 1;
    }
}
return -1;

medium

A. -1

B. 6

C. 0

D. 4

Solution

1. Step 1: Understand the array and target

   The array is rotated: [4,5,6,7,0,1,2]. Target is 3, which is not in the array.

2. Step 2: Trace the binary search steps

   Binary search checks mid and adjusts left/right. Since 3 is not present, loop ends and returns -1.

3. Final Answer:

   -1 -> Option A

4. Quick Check:

   Target missing returns -1 [OK]

Hint: If target not found, binary search returns -1 [OK]

Common Mistakes:

• Assuming target 3 is in array
• Returning wrong index for missing target
• Not updating left/right correctly

4. Identify the bug in this C# code snippet for searching target in rotated sorted array:

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

medium

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

B. Mid calculation should be mid = left + (right - left) / 2

C. The target comparison operators are incorrect

D. The return -1 should be inside the loop

Solution

1. Step 1: Analyze loop condition

   The loop uses left < right, which misses the case when left == right, possibly skipping the last element check.

2. Step 2: Correct loop condition for binary search

   Binary search requires left <= right to ensure all elements are checked, including when left equals right.

3. Final Answer:

   The loop condition should be left <= right, not left < right -> Option A

4. Quick Check:

   Use left <= right for complete search [OK]

Hint: Use <= in loop condition to check all elements [OK]

Common Mistakes:

• Using < instead of <= in loop
• Incorrect mid calculation causing overflow
• Misplaced return statements

5. Given a rotated sorted array int[] nums = {15, 18, 2, 3, 6, 12};, write a C# method to find the index of target 3 using binary search in rotated array. What is the returned index?

hard

A. 2

B. 3

C. 4

D. -1

Solution

1. Step 1: Understand the array and target

   The array is rotated: [15, 18, 2, 3, 6, 12]. Target is 3, which is present at index 3.

2. Step 2: Apply binary search in rotated array

   Using the binary search logic for rotated arrays, the search will find target 3 at index 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Target 3 found at index 3 [OK]

Hint: Trace mid and adjust left/right to find target index [OK]

Common Mistakes:

• Confusing index of 3 with nearby elements
• Not handling rotation correctly
• Returning -1 for existing target