Search in Rotated Sorted Array in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of search in rotated sorted array

What is the output of the following C# code that searches for the target 3 in a rotated sorted array?

DSA C#

using System;

class Program {
    static int Search(int[] nums, int target) {
        int left = 0, right = nums.Length - 1;
        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;
    }

    static void Main() {
        int[] arr = {4,5,6,7,0,1,2};
        Console.WriteLine(Search(arr, 3));
    }
}

A -1

B 4

C 2

D 0

💡 Hint

Think about whether the target 3 exists in the array.

The array is rotated and does not contain the number 3. The search returns -1 when the target is not found.

❓ Predict Output

intermediate

Find index of target in rotated sorted array

What index does the following code return when searching for target 0 in the rotated sorted array?

DSA C#

using System;

class Program {
    static int Search(int[] nums, int target) {
        int left = 0, right = nums.Length - 1;
        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;
    }

    static void Main() {
        int[] arr = {4,5,6,7,0,1,2};
        Console.WriteLine(Search(arr, 0));
    }
}

A 3

B 0

C 4

D -1

💡 Hint

Check the position of 0 in the array.

The target 0 is at index 4 in the array.

🔧 Debug

advanced

Identify the error in rotated array search code

What error does the following code produce when searching for target 5 in the rotated sorted array?

DSA C#

using System;

class Program {
    static int Search(int[] nums, int target) {
        int left = 0, right = nums.Length - 1;
        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;
    }

    static void Main() {
        int[] arr = {6,7,0,1,2,4,5};
        Console.WriteLine(Search(arr, 5));
    }
}

A Returns -1 even though 5 is in the array

B Throws IndexOutOfRangeException

C Infinite loop

D Returns 0 always

💡 Hint

Check the loop condition and how it affects the search.

The loop uses 'left < right' which can skip checking the last element, causing the function to miss the target and return -1.

❓ Predict Output

advanced

Output of search for target 1 in rotated array

What is the output of the following code searching for target 1 in the rotated sorted array?

DSA C#

using System;

class Program {
    static int Search(int[] nums, int target) {
        int left = 0, right = nums.Length - 1;
        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;
    }

    static void Main() {
        int[] arr = {5,6,7,8,9,1,2,3,4};
        Console.WriteLine(Search(arr, 1));
    }
}

A 0

B 5

C -1

D 4

💡 Hint

Locate the index of 1 in the array.

The target 1 is at index 5 in the array.

🧠 Conceptual

expert

Minimum number of comparisons in rotated sorted array search

What is the minimum number of comparisons needed in the worst case to find a target in a rotated sorted array of size n using binary search?

A O(n log n)

B O(n)

C O(1)

D O(log n)

💡 Hint

Think about how binary search divides the search space.

Binary search on a rotated sorted array still divides the search space in half each step, so it takes O(log n) comparisons in the worst case.

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