Search 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 point. 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 use simple binary search directly on a rotated sorted array?

Because the array is not fully sorted in ascending order due to rotation, simple binary search may fail. The array has two sorted parts separated by the pivot, so we need to adjust the search to handle this.

intermediate

In searching a rotated sorted array, how do we decide which half to search next?

We check which half is sorted by comparing the middle element with the start element. If the left half is sorted and the target lies within it, we search left; otherwise, we search right. If the right half is sorted and the target lies within it, we search right; otherwise, left.

beginner

What is the time complexity of searching in a rotated sorted array using modified binary search?

The time complexity is O(log n) because we still halve the search space each step, just like regular binary search, but with extra checks for rotation.

beginner

Show the base case for the recursive search in a rotated sorted array.

If the search range is invalid (start > end), return -1 to indicate the target is not found.

What does a rotated sorted array look like?

A [1,3,2,4,6,5,7]

B [1,2,3,4,5,6,7]

C [7,6,5,4,3,2,1]

D [4,5,6,7,0,1,2]

Option D is a sorted array rotated at pivot 3. Others are fully sorted or unsorted.

If the middle element is greater than the start element, which half is sorted?

A Left half

B Right half

C Both halves

D None

If middle > start, left half is sorted.

What is the first step in searching a rotated sorted array?

A Find the pivot point

B Compare target with middle element

C Check if array is empty

D Sort the array

We start by comparing target with middle element to decide search direction.

What is the worst-case time complexity of searching in a rotated sorted array?

A O(n)

B O(1)

C O(log n)

D O(n log n)

Modified binary search keeps time complexity at O(log n).

If the target is not found in the array, what should the search function return?

A -1

B 0

C null

D Index of closest element

Returning -1 indicates target not found.

Explain step-by-step how to search for a target in a rotated sorted array using binary search.

Think about how rotation splits the array into two sorted parts.

Describe why a rotated sorted array requires a modified binary search instead of a normal binary search.

Consider what breaks the normal binary search assumptions.

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